Machine Learning & Artificial Intelligence

Prompt:
# Hybrid Efficiency Block — Forward Pass Computation

A hybrid efficiency block processes $T = 4$ tokens in a $d = 8$-dimensional embedding space. The input $X \in \mathbb{R}^{4 \times 8}$ is sampled from `numpy.random.RandomState(37).randn(4, 8)`.

The block integrates four components. The input $X$ is first normalized with RMSNorm producing $x_n$. The normalized $x_n$ feeds **only** the retention branch. The raw input $X$ feeds both the selective state-space branch and the mixture-of-experts branch.

**Numerical conventions:** All computations must be performed in float64 using NumPy semantics. Matrix multiplications follow standard row-by-column rules with explicit broadcasting as defined. Softmax is applied along the last axis unless otherwise noted. The sigmoid function is $\sigma(x) = 1/(1 + e^{-x})$. All intermediate values must be computed without rounding until the final answers, which are rounded to 4 decimal places.

---

## Component 1: RMSNorm

$$x_n[i, :] = \frac{X[i, :]}{\sqrt{\frac{1}{d}\sum_{j=1}^{d} X[i,j]^2 + \varepsilon}} \quad \text{where } d = 8,\; \varepsilon = 10^{-5}$$

No mean subtraction. Gain vector is $\mathbf{1}_8$.

---

## Component 2: Selective State-Space Model (on raw $X$)

A one-dimensional selective scan with continuous parameter $A = -1$. For each token $t = 0, \ldots, 3$:

1. **Projections:**
$$B_t = X_t \cdot W_B \in \mathbb{R}^{8}, \quad C_t = X_t \cdot W_C \in \mathbb{R}^{8}, \quad \Delta_t = \mathrm{softplus}(X_t \cdot W_\Delta) \in \mathbb{R}$$
where $\mathrm{softplus}(x) = \ln(1 + e^x)$ and $W_\Delta \in \mathbb{R}^{8 \times 1}$ so $\Delta_t$ is a scalar.

2. **Zero-order hold discretization:**
$$\bar{A}_t = e^{\Delta_t \cdot A} = e^{-\Delta_t} \in \mathbb{R}, \quad \bar{B}_t = (1 - e^{-\Delta_t}) \cdot B_t \in \mathbb{R}^{8}$$

3. **Recurrence** ($h_0 = \mathbf{0} \in \mathbb{R}^{8}$):
$$h_t = \bar{A}_t \cdot h_{t-1} + \bar{B}_t \in \mathbb{R}^{8}$$
where $\bar{A}_t$ is a scalar broadcasting across all 8 dimensions.

4. **Output** (element-wise product):
$$y_{\text{ssm}}[t] = C_t \odot h_t \in \mathbb{R}^{8}$$

---

## Component 3: Retention Mechanism (on $x_n$)

With $n_{\text{heads}} = 2$, $d_{\text{head}} = 4$, decay $\gamma = 0.9$.

1. **Projections:**
$$Q = x_n \cdot W_{Q_r} \in \mathbb{R}^{4 \times 8}, \quad K = x_n \cdot W_{K_r} \in \mathbb{R}^{4 \times 8}, \quad V = x_n \cdot W_{V_r} \in \mathbb{R}^{4 \times 8}$$

2. **Reshape** to $Q^{(h)}, K^{(h)}, V^{(h)} \in \mathbb{R}^{4 \times 4}$ for each head $h \in \{0, 1\}$, where head $h$ takes columns $[4h : 4h+4]$.

3. **Causal decay matrix** $D \in \mathbb{R}^{4 \times 4}$:
$$D_{ij} = \begin{cases} \gamma^{i-j} & \text{if } i \geq j \\ 0 & \text{otherwise} \end{cases}$$

4. **Retention scores and output per head:**
$$S^{(h)} = Q^{(h)} (K^{(h)})^\top \in \mathbb{R}^{4 \times 4}, \quad R^{(h)} = (S^{(h)} \odot D) \cdot V^{(h)} \in \mathbb{R}^{4 \times 4}$$

5. **Concatenate** heads and project:
$$y_{\text{ret}} = [R^{(0)} \| R^{(1)}] \cdot W_{O_r} \in \mathbb{R}^{4 \times 8}$$

6. **Retention entropy** $H_{\text{ret}}$: For each head $h$ and each query position $i$:
   - Compute decay-weighted scores: $z^{(h)}_{ij} = S^{(h)}_{ij} \cdot D_{ij}$ for $j \leq i$. Set $z^{(h)}_{ij} = -\infty$ for $j > i$ (causal mask).
   - Row-normalize via softmax: $W^{(h)}_{ij} = \frac{\exp(z^{(h)}_{ij})}{\sum_{k \leq i} \exp(z^{(h)}_{ik})}$
   - Entropy per position: $H^{(h)}_i = -\sum_{j \leq i} W^{(h)}_{ij} \ln W^{(h)}_{ij}$
   - Average: $H_{\text{ret}} = \frac{1}{n_{\text{heads}} \cdot T} \sum_{h,i} H^{(h)}_i$ where $T = 4$.

---

## Component 4: Sparse Mixture of Experts (on raw $X$)

With 2 experts.

1. **Router:** $\text{logits} = X \cdot W_{\text{router}} \in \mathbb{R}^{4 \times 2}$
2. **Routing probabilities:** $\text{probs} = \mathrm{softmax}(\text{logits}, \text{axis}=-1) \in \mathbb{R}^{4 \times 2}$
3. **Top-2 selection:** Since there are exactly 2 experts, both are always selected and weights equal the original softmax probabilities (no renormalization needed).
4. **Expert outputs and combination:**
$$y_{\text{moe}}[t] = \text{probs}[t, 0] \cdot (X_t \cdot W_{e_1}) + \text{probs}[t, 1] \cdot (X_t \cdot W_{e_2}) \in \mathbb{R}^{8}$$

---

## Component 5: Linear Attention Fusion

1. **Feature map** $\phi$ (ELU+1), applied element-wise:
$$\phi(x) = \begin{cases} x + 1 & \text{if } x \geq 0 \\ e^x & \text{if } x < 0 \end{cases}$$

2. **Sources:** Query from $y_{\text{ret}}$, Key and Value from $y_{\text{ssm}}$.

3. **Causal linear attention** for each position $i = 0, \ldots, 3$:
$$h_{\text{lin}}[i] = \frac{\sum_{j=0}^{i} \bigl(\phi(Q_i) \cdot \phi(K_j)\bigr) \cdot V_j}{\sum_{j=0}^{i} \phi(Q_i) \cdot \phi(K_j)}$$
where $Q_i = y_{\text{ret}}[i] \in \mathbb{R}^8$, $K_j = y_{\text{ssm}}[j] \in \mathbb{R}^8$, $V_j = y_{\text{ssm}}[j] \in \mathbb{R}^8$, and $\cdot$ denotes the dot product (scalar). The numerator is a weighted sum of $V_j$ vectors; the denominator is a scalar sum of dot products.

4. **Gated residual:**
$$h_{\text{base}} = \frac{y_{\text{ssm}} + y_{\text{ret}} + y_{\text{moe}}}{3}$$
$$\text{gate} = \sigma(h_{\text{base}} \cdot W_r) \quad \text{where } \sigma(x) = \frac{1}{1 + e^{-x}}$$
$$h = h_{\text{base}} + \text{gate} \odot h_{\text{lin}}$$

---

## Component 6: SwiGLU Feed-Forward Network

With $d_{\text{ff}} = 12$ and $\mathrm{Swish}(z) = z \cdot \sigma(z) = \frac{z}{1 + e^{-z}}$.

1. $h_n = \mathrm{RMSNorm}(h)$ (same formula as Component 1, applied to $h$).
2. $\text{gate\_val} = \mathrm{Swish}(h_n \cdot W) \in \mathbb{R}^{4 \times 12}$, $\quad \text{linear\_val} = h_n \cdot V \in \mathbb{R}^{4 \times 12}$
3. $\mathrm{SwiGLU}(h_n) = (\text{gate\_val} \odot \text{linear\_val}) \cdot W_{\text{down}} \in \mathbb{R}^{4 \times 8}$
4. Residual: $h_{\text{final}} = h + \mathrm{SwiGLU}(h_n) \in \mathbb{R}^{4 \times 8}$

---

## Task Heads

### Classification (Focal Loss)
- Logits: $\ell = h_{\text{final}} \cdot W_{\text{cls}} \in \mathbb{R}^{4 \times 4}$
- Class probabilities: $p = \mathrm{softmax}(\ell, \text{axis}=-1)$
- Label smoothing ($\varepsilon = 0.1$, $K = 4$ classes, labels $= [2, 0, 3, 1]$):
$$y_s[t, k] = \begin{cases} 1 - \varepsilon + \varepsilon/K & \text{if } k = \text{label}[t] \\ \varepsilon / K & \text{otherwise} \end{cases}$$
- Per-token cross-entropy: $\text{CE}[t] = -\sum_{k} y_s[t,k] \cdot \ln p[t,k]$
- Per-token focal weight using **unsmoothed** true class probability: $p_t = p[t, \text{label}[t]]$
- Focal loss: $L_{\text{cls}} = \frac{1}{T} \sum_t (1 - p_t)^\gamma \cdot \text{CE}[t]$ with $\gamma = 2$

### Regression (Huber Loss)
- Predictions: $\hat{y} = h_{\text{final}} \cdot W_{\text{reg}} \in \mathbb{R}^{4 \times 2}$
- Targets: $y = \text{RandomState}(71).\text{randn}(4, 2)$
- Element-wise error: $e_{ij} = y_{ij} - \hat{y}_{ij}$
$$\text{Huber}(e) = \begin{cases} \frac{1}{2} e^2 & \text{if } |e| \leq \delta \\ \delta(|e| - \frac{1}{2}\delta) & \text{otherwise} \end{cases} \quad \text{with } \delta = 1.0$$
- $L_{\text{reg}} = \frac{1}{4 \cdot 2} \sum_{i,j} \text{Huber}(e_{ij})$

### KL Divergence
- Token-averaged class distribution: $\bar{p} = \frac{1}{T} \sum_t p[t, :]$
- $D_{\text{KL}} = \sum_{k=1}^{K} \bar{p}_k \ln(K \cdot \bar{p}_k)$

---

## Total Loss

$$H = H_{\text{ret}}$$
$$E = L_{\text{cls}} + 0.5 \cdot L_{\text{reg}} + 0.01 \cdot H + 0.1 \cdot D_{\text{KL}}$$

---

## Weight Initialization

All 16 weight matrices drawn from a single `RandomState(97)` in the order below, each as `rng.randn(rows, cols) * std`:

| # | Weight | Shape | std | Role |
|---|--------|-------|-----|------|
| 1 | $W_B$ | $8 \times 8$ | $1/\sqrt{8}$ | SSM B projection |
| 2 | $W_C$ | $8 \times 8$ | $1/\sqrt{8}$ | SSM C projection |
| 3 | $W_\Delta$ | $8 \times 1$ | $1/\sqrt{8}$ | SSM step size |
| 4 | $W_{Q_r}$ | $8 \times 8$ | $1/\sqrt{8}$ | Retention Q |
| 5 | $W_{K_r}$ | $8 \times 8$ | $1/\sqrt{8}$ | Retention K |
| 6 | $W_{V_r}$ | $8 \times 8$ | $1/\sqrt{8}$ | Retention V |
| 7 | $W_{O_r}$ | $8 \times 8$ | $1/\sqrt{8}$ | Retention output |
| 8 | $W_{\text{router}}$ | $8 \times 2$ | $1/\sqrt{8}$ | MoE router |
| 9 | $W_{e_1}$ | $8 \times 8$ | $1/\sqrt{8}$ | MoE expert 1 |
| 10 | $W_{e_2}$ | $8 \times 8$ | $1/\sqrt{8}$ | MoE expert 2 |
| 11 | $W_r$ | $8 \times 8$ | $1/\sqrt{8}$ | Gated residual |
| 12 | $V$ | $8 \times 12$ | $1/\sqrt{8}$ | SwiGLU linear |
| 13 | $W$ | $8 \times 12$ | $1/\sqrt{8}$ | SwiGLU gate |
| 14 | $W_{\text{down}}$ | $12 \times 8$ | $1/\sqrt{12}$ | SwiGLU down |
| 15 | $W_{\text{cls}}$ | $8 \times 4$ | $1/\sqrt{8}$ | Classification |
| 16 | $W_{\text{reg}}$ | $8 \times 2$ | $1/\sqrt{8}$ | Regression |

All biases initialized to zero.

---

## Questions

**Q1:** What is $H$ (retention entropy)? Round to 4 decimal places.

**Q2:** What is $L_{\text{cls}}$ (focal loss)? Round to 4 decimal places.

**Q3:** What is $L_{\text{reg}}$ (Huber loss)? Round to 4 decimal places.

**Q4:** What is $D_{\text{KL}}$? Round to 4 decimal places.

**Q5:** What is $E$ (total loss)? Round to 4 decimal places.

Does the answer compute x_n[0,0] = -0.0638 and x_n[1,3] = 1.8106 from the RMSNorm step (values rounded to 4 decimal places)?

Does the answer compute y_ssm[0,0] = -0.3166 and y_ssm[2,3] = -0.1890 from the selective SSM recurrence (values rounded to 4 decimal places)?

Does the answer compute y_ret[0,0] = -0.3135 and y_ret[2,3] = 1.1332 from the retention mechanism (values rounded to 4 decimal places)?

Does the answer report H_ret = 0.2587 (rounded to 4 decimal places)?

Does the answer compute h_final[0,0] = 2.2156 and h_final[2,3] = 0.5309 after the SwiGLU residual (values rounded to 4 decimal places)?

Does the answer report L_cls (focal loss) = 2.1703 (rounded to 4 decimal places)?

Does the answer report L_reg (Huber loss) = 1.1017 (rounded to 4 decimal places)?

Does the answer report D_KL = 0.1544 and E (total loss) = 2.7392 (rounded to 4 decimal places)?

Does the answer include checksums sum(x_n) = 6.3585, sum(y_ssm) = -0.3665, sum(y_ret) = -0.3220, and sum(h_final) = 13.0751 (rounded to 4 decimal places)?


Prompt:
# Distributed Operating Systems - Coordinated Execution Analysis (Deterministic Trace, Derived-Constants Spec)

## 0) Reading rules (AUTHORITATIVE)

This document contains:
- AUTHORITATIVE rules (must be followed)
- LEGACY notes (included for realism; must be ignored if they conflict)

If any statement conflicts, AUTHORITATIVE wins.

You must reconstruct a deterministic execution trace over:
- t in [0,50)

All outputs must be derived only from this prompt. No unstated behaviors are allowed.

## 1) Canonical constants (AUTHORITATIVE)

Let:
- Q = 10 (ms)   (the time quantum)
- HALFQ = Q/2 = 5 (ms)
- ONE = 1 (ms)
- TWO = 2 (ms)

All times in this task are integers in milliseconds.

Define these derived timestamps (all computed from Q, HALFQ, ONE, TWO):
- t_P3_spawn = Q + TWO
- t_gc_read = 2*Q + (Q - TWO)     (note: Q - TWO = 8)
- t_pf_start = t_gc_read
- t_pf_end = t_pf_start + Q
- t_timeout = 4*Q + (HALFQ)       (note: HALFQ=5)
- t_sleep_end = t_timeout + ONE
- t_cleanup_end = t_sleep_end + TWO

Define the three P1 store times for V:
- t_V1 = Q
- t_V2 = 2*Q
- t_V3 = 3*Q

Define the 2PC phase boundaries:
- t_prepare_send = 0
- t_commit_send = t_V3

## 2) Time ordering (AUTHORITATIVE)

If multiple events occur at the same timestamp t, process them in this exact order:
1) Network deliveries scheduled at time t
2) Unblocks whose unblock-time is exactly t
3) Scheduler decisions at time t (preemption/dispatch)
4) CPU-defined memory operations specified to occur exactly at time t

Snapshots at time t must be taken after steps (1)-(4).

Cross-node linearization for CPU-defined memory operations at the same timestamp t:
- Node A operations happen first, then Node B, then Node C
All coherence reactions complete immediately before the next linearized operation.

LEGACY note (ignore): "Snapshots are taken before CPU ops." (This LEGACY note is false for this task.)

## 3) Scheduler (AUTHORITATIVE)

- Preemptive fixed priority scheduling: higher numeric priority preempts immediately.
- Exact quantum: Q ms per dispatch.
- If a process blocks or is preempted, remaining quantum is discarded.
- If a process continues past a quantum boundary while still running, treat it as a fresh dispatch with a new Q ms quantum.
- Context switch cost is 0ms.
- Message handling cost is 0ms and does not consume user CPU time.

## 4) Shared mapped page F (AUTHORITATIVE)

There is exactly one scored shared memory-mapped page F with two 32-bit words:
- L on cache line CL_L (lock word)
- V on cache line CL_V (data word)

Initial state at t=0-:
- L = 0
- V = 0
- MESI states:
  - CL_L: A=S, B=S, C=S
  - CL_V: A=S, B=S, C=S

Only CL_L and CL_V are scored for MESI output.

## 5) Lock semantics on L (AUTHORITATIVE)

Encoding:
- L = 0 unlocked
- L = 1 writer lock held by P1
- L = 2 writer lock held by P2

Writer preference:
- If any writer is waiting (spinning), all reader acquisitions fail until a writer acquires and releases.

Deterministic spin algorithm (runs only while scheduled):
- At each integer ms boundary while scheduled, a waiting process performs exactly one non-temporal load of L.
- Non-temporal loads do not allocate CL_L and do not change the local MESI state of CL_L.
- A writer performs CAS(L,0,writer_id) only if its immediately preceding non-temporal load observed L==0.
- A reader acquires only if its load observed L==0 AND there are no waiting writers.

LEGACY note (ignore): "Spin loads allocate cache lines like normal reads." (This LEGACY note is false for this task.)

## 6) Processes and priorities (AUTHORITATIVE)

### Node A
- P1 priority = 2, runnable at t=0, only user process on Node A.
- At t=0, P1 acquires writer lock via cached store L := 1.

P1 performs three updates while holding the lock:
- U1 compute over [0,t_V1), then at t_V1 do cached store V := V + 1
- U2 compute over [t_V1,t_V2), then at t_V2 do cached store V := V + 1
- U3 compute over [t_V2,t_V3), then at t_V3 do cached store V := V + 1

2PC coordinator actions:
- At t_prepare_send, send PREPARE_RESERVE(T).
- At t_commit_send, after the V store, send COMMIT(T,V_final) and then block waiting for ACKs.

Blocking semantics on Node A:
- When P1 blocks, it is not runnable and consumes 0 CPU.
- If no other runnable processes exist (true here), CPU is idle.
- In CPU traces, BLK means "P1 blocked and CPU idle".

Lock release actions at the unique release time t_release (defined by the 2PC release rule below):
At timestamp t_release, Node A performs in this order:
1) explicit write-back of CL_V and downgrade CL_V M->E if no sharers exist at that instant
2) cached store L := 0 (unlock)

After releasing, P1 runs CPU-only cleanup over [t_release,50).

### Node B
- P2 priority = 2, runnable at t=0; spins for writer lock when scheduled; never blocks in [0,50).
- P3 priority = 3, becomes runnable at t_P3_spawn.
  - From t_P3_spawn until t_timeout, attempts read lock (always fails in this run).
  - At t_pf_start while running, touches metadata and blocks due to a page fault for [t_pf_start,t_pf_end).
  - At t_timeout, calls sleep(ONE), blocking [t_timeout,t_sleep_end).
  - Over [t_sleep_end,t_cleanup_end), runs CPU-only cleanup for exactly TWO ms, then blocks [t_cleanup_end,50).

Deterministic post-release access:
- At timestamp t_release, after Node A has executed the unlock store L:=0 in the cross-node linearization, P2 performs a successful acquisition CAS and then cached store L := 2 at timestamp t_release.
- At timestamp (t_release + ONE), P2 performs exactly one cached read PrRd(V) and performs no writes to V in [t_release,50).

### Node C
- GC priority = 1, runnable at t=0, runs continuously over [0,4*Q).
- At t_gc_read exactly, GC performs one cached read PrRd(V).
- Node C is idle over [4*Q,50). Use IDLE in traces.

## 7) 2PC protocol constants (AUTHORITATIVE)

Coordinator: Node A (P1)
Participants: Node B, Node C

Network bounds: one-way latency is within [5,15] ms.
For this deterministic run, latencies are defined from Q and HALFQ:

- PREPARE_RESERVE latency A->B = Q, and A->C = Q
- VOTE_YES latency B->A = Q, and C->A = Q
- COMMIT latency A->B = HALFQ, and A->C = HALFQ
- ACK_COMMIT latency B->A = Q, and C->A = HALFQ

Prepare phase:
- t_prepare_send: A sends PREPARE_RESERVE(T)
- t_prepare_send + Q: B and C receive PREPARE, log prepared instantly, send VOTE_YES(T)
- t_prepare_send + 2*Q: A receives both VOTE_YES(T)

Commit phase:
- t_commit_send: A sends COMMIT(T,V_final)
- t_commit_send + HALFQ: B and C receive COMMIT
- On receipt of COMMIT, each participant must touch its commit-log page before sending ACK_COMMIT.
- Node B commit-log page fault rule: Node B cannot touch its commit-log page during [t_pf_start,t_pf_end). If COMMIT arrives during that interval, B delays its ACK until t_pf_end.
- Node C has no delay.

Coordinator release rule (defines t_release uniquely):
- P1 releases the writer lock only after receiving ACK_COMMIT from both B and C.

## 8) MESI-like coherence model (AUTHORITATIVE)

States: M, E, S, I.

- Cached store: writer obtains ownership, ends in M, invalidates other nodes to I.
- Cached read PrRd on I: requester becomes S.
- If another node holds M and a cached read occurs elsewhere: owner flushes and downgrades M->S, requester becomes S.
- At t_release, Node A explicitly write-backs CL_V and downgrades M->E if no other sharers exist at that instant.
- At t_release + ONE, Node B cached reads V once, causing A: E->S and B: I->S for CL_V.
- Non-temporal spin loads of L do not allocate CL_L and do not change CL_L MESI states.

## 9) Tasks

Compute deterministic outputs for t in [0,50):

1) Timeline for exactly these 12 intervals:
   [0,10), [10,12), [12,20), [20,28), [28,30), [30,35), [35,38), [38,40), [40,45), [45,46), [46,48), [48,50)

2) Identify deadlock, livelock, and priority inversion points using explicit times.

3) Final state:
- final L and V at t=50-
- MESI snapshots for CL_L and CL_V for all nodes at t = 0, 10, 35, 48, 50
- mention the transient coherence event at t=28 caused by GC's cached read of V

4) Numeric performance:
- throughput = committed updates per second over the 50ms window
- lock utilization efficiency:
  \[
  \text{efficiency} = \frac{\text{productive lock time}}{\text{lock held time}}
  \]
  productive lock time is 30ms, lock held time is from t=0 to t_release

## 10) Required Output Lines (Keys and Formats)

Your final answer must include these keys exactly once each, using these formats:

SNAPSHOT_TIMES_MS = [0,10,35,48,50]

INTERVALS = [[0,10),[10,12),[12,20),[20,28),[28,30),[30,35),[35,38),[38,40),[40,45),[45,46),[46,48),[48,50)]

CPU_TRACES = [A:[x1,...,x12],B:[y1,...,y12],C:[z1,...,z12]]
- Each xi, yi, zi is one of: P1, P2, P3, GC, BLK, IDLE

V_FINAL = <integer>
L_FINAL = <integer>

MESI_CL_V_ALL = [A:[s0,s10,s35,s48,s50],B:[t0,t10,t35,t48,t50],C:[u0,u10,u35,u48,u50],T28:A:p->q,C:r->w]

MESI_CL_L_ALL = [A:[a0,a10,a35,a48,a50],B:[b0,b10,b35,b48,b50],C:[c0,c10,c35,c48,c50]]

THROUGHPUT_UPDATES_PER_SEC = <integer>
LOCK_EFFICIENCY_PERCENT = <decimal>

PRIORITY_INVERSION_DURATION_MS = <integer>
LIVELOCK_PRESENT = <YES/NO>
DEADLOCK_POSSIBLE_2PC_BLOCKING = <YES/NO>

Does the answer contain the exact line 'SNAPSHOT_TIMES_MS = [0,10,35,48,50]'?

Does the answer contain the exact line 'CPU_TRACES = [A:[P1,P1,P1,P1,P1,BLK,BLK,BLK,BLK,BLK,BLK,P1],B:[P2,P2,P3,P3,P2,P2,P2,P3,P3,P2,P3,P2],C:[GC,GC,GC,GC,GC,GC,GC,GC,IDLE,IDLE,IDLE,IDLE]]'?

Does the answer contain the exact line 'V_FINAL = 3'?

Does the answer contain the exact line 'L_FINAL = 2'?

Does the answer contain the exact line 'MESI_CL_V_ALL = [A:[S,M,M,E,S],B:[S,I,I,I,S],C:[S,I,I,I,I],T28:A:M->S,C:I->S]'?

Does the answer contain the exact line 'MESI_CL_L_ALL = [A:[M,M,M,I,I],B:[I,I,I,M,M],C:[I,I,I,I,I]]'?

Does the answer contain the exact line 'THROUGHPUT_UPDATES_PER_SEC = 60'?

Does the answer contain the exact line 'LOCK_EFFICIENCY_PERCENT = 62.5'?

Does the answer contain the exact line 'PRIORITY_INVERSION_DURATION_MS = 33'?

Does the answer contain the exact line 'LIVELOCK_PRESENT = YES'?

Does the answer contain the exact line 'DEADLOCK_POSSIBLE_2PC_BLOCKING = YES'?

Does the answer explicitly state that t_release = 48 derived from receiving both ACK_COMMIT messages after the Node B delay due to the page fault interval [28,38)?

Does the answer correctly explain the transient MESI event at t=28 where GC's cached read forces A: M->S and C: I->S for CL_V?

Does the answer correctly reflect that Node B's ACK is delayed until t_pf_end = 38 due to the page fault interval, leading to final ACK arrival at t=48?

## 1) Compute derived timestamps from Q=10 and HALFQ=5

Q=10, HALFQ=5, ONE=1, TWO=2

t_P3_spawn = Q + TWO = 12
t_gc_read = 2*Q + (Q - TWO) = 20 + 8 = 28
t_pf_start = 28
t_pf_end = t_pf_start + Q = 28 + 10 = 38
t_timeout = 4*Q + HALFQ = 40 + 5 = 45
t_sleep_end = t_timeout + ONE = 46
t_cleanup_end = t_sleep_end + TWO = 48

t_V1 = Q = 10
t_V2 = 2*Q = 20
t_V3 = 3*Q = 30

## 2) Determine t_release from 2PC rules

Prepare:
- t_prepare_send=0
- PREPARE arrives at 0+Q=10
- VOTE_YES arrives at 0+2Q=20

Commit:
- t_commit_send = t_V3 = 30
- COMMIT arrives at 30+HALFQ = 35

ACKs:
- Node C ACK latency is HALFQ, and Node C has no log delay:
  - C sends ACK at 35, A receives at 35+HALFQ = 40
- Node B cannot touch commit log during [t_pf_start,t_pf_end) = [28,38)
  - COMMIT arrives at 35 within [28,38), so B delays sending ACK until 38
  - B->A ACK latency is Q, so A receives B ACK at 38+Q = 48

Coordinator releases only after both ACKs received, so:
t_release = 48

## 3) Timeline (t in [0,50))

| Interval | Node A | Node B | Node C |
|---|---|---|---|
| [0,10)  | P1  | P2  | GC   |
| [10,12) | P1  | P2  | GC   |
| [12,20) | P1  | P3  | GC   |
| [20,28) | P1  | P3  | GC   |
| [28,30) | P1  | P2  | GC   |
| [30,35) | BLK | P2  | GC   |
| [35,38) | BLK | P2  | GC   |
| [38,40) | BLK | P3  | GC   |
| [40,45) | BLK | P3  | IDLE |
| [45,46) | BLK | P2  | IDLE |
| [46,48) | BLK | P3  | IDLE |
| [48,50) | P1  | P2  | IDLE |

## 4) Deadlock, Livelock, Priority Inversion

Priority inversion:
- P3 attempts from t=12 until t=45 but cannot acquire while P1 holds writer lock until t=48 and writer preference blocks readers while writer P2 is waiting.
- Duration = 45 - 12 = 33ms.

Livelock:
- P2 and P3 repeatedly retry lock acquisition while L remains nonzero until t=48, consuming CPU without acquisition progress until release.

Deadlock possible (2PC blocking):
- 2PC is a blocking protocol if the coordinator fails after participants have voted YES; in this run there is no failure but blocking risk exists by protocol structure.

## 5) Final values and MESI snapshots

Final values at t=50-:
- V_FINAL = 3
- L_FINAL = 2

CL_V transient at t=28:
- At t=28, A holds CL_V in M (from store at t=20).
- GC cached PrRd(V) forces A: M->S and C: I->S.
- At t=30, P1 stores V:=3, forcing A: S->M and C: S->I.

CL_V snapshots at t = 0,10,35,48,50:
- t=0: A=S, B=S, C=S
- t=10: A=M, B=I, C=I
- t=35: A=M, B=I, C=I
- t=48: A=E, B=I, C=I
- t=50: A=S, B=S, C=I

CL_L snapshots at t = 0,10,35,48,50:
- t=0: A=M, B=I, C=I
- t=10: A=M, B=I, C=I
- t=35: A=M, B=I, C=I
- t=48: A=I, B=M, C=I
- t=50: A=I, B=M, C=I

## 6) Performance

Throughput:
- 3 committed updates over 50ms = 0.050s
- Throughput = 3 / 0.050 = 60 updates/s

Lock utilization efficiency:
- Productive lock time = 30ms
- Lock held time = t_release - 0 = 48ms
\[
\text{efficiency} = \frac{30}{48} = 0.625 = 62.5
\]

## 7) Required Output Lines

SNAPSHOT_TIMES_MS = [0,10,35,48,50]

INTERVALS = [[0,10),[10,12),[12,20),[20,28),[28,30),[30,35),[35,38),[38,40),[40,45),[45,46),[46,48),[48,50)]

CPU_TRACES = [A:[P1,P1,P1,P1,P1,BLK,BLK,BLK,BLK,BLK,BLK,P1],B:[P2,P2,P3,P3,P2,P2,P2,P3,P3,P2,P3,P2],C:[GC,GC,GC,GC,GC,GC,GC,GC,IDLE,IDLE,IDLE,IDLE]]

V_FINAL = 3
L_FINAL = 2

MESI_CL_V_ALL = [A:[S,M,M,E,S],B:[S,I,I,I,S],C:[S,I,I,I,I],T28:A:M->S,C:I->S]

MESI_CL_L_ALL = [A:[M,M,M,I,I],B:[I,I,I,M,M],C:[I,I,I,I,I]]

THROUGHPUT_UPDATES_PER_SEC = 60
LOCK_EFFICIENCY_PERCENT = 62.5

PRIORITY_INVERSION_DURATION_MS = 33
LIVELOCK_PRESENT = YES
DEADLOCK_POSSIBLE_2PC_BLOCKING = YES


Prompt:
## Problem

You are given an `n × m` grid (`1 ≤ n,m ≤ 200`) filled with digits `0..9`.

For each digit `d` (`0..9`), you are also given a move vector `(a[d], b[d])` (each component in `[-200,200]`).

Define Wilbur’s movement rule:

- If Wilbur is currently at cell `(x,y)` and the digit there is `d`, then his **forced next position** is `(x + a[d], y + b[d])` **if** that cell is inside the grid.
- If the forced next position is outside the grid, Wilbur **stays** at `(x,y)`.

Before each forced move, Wilbur may either:
- append the current cell’s digit to a whiteboard string, or
- append nothing.

Digits appended over time form a string (appended to the end each time).

You are given `q` query strings `s1..sq` (digits only). For each query `s`, determine whether there exists **some starting cell** such that, after finitely many forced moves (and optional writes), Wilbur can obtain **exactly** the string `s` on the whiteboard.

### Input
- `n m q` (`1..200`)
- `n` lines of `m` digits each (the grid)
- 10 lines: for digit `d`, line contains `a[d] b[d]`
- `q` lines: the query strings (total length over all queries ≤ `1,000,000`)

### Output
For each query, print `YES` if possible, else `NO`.

The answer models the forced movement as a directed graph on the grid cells in which every vertex has outdegree exactly 1.

The answer states that when the forced next position is outside the grid, the transition is a self-loop (the cell transitions to itself).

The answer states that any writable whiteboard string is a subsequence of the digit sequence along the forced walk from a chosen start cell.

The answer states that each (weakly) connected component of a functional graph contains exactly one directed cycle.

The answer defines the reverse adjacency relation in the form rev[u] = { v : to[v] = u}?

The answer states that the query string is reversed (r = reverse(s)) for the main search procedure.

The answer defines a per-query transition nxt(pos, d) that advances the matched position by 1 exactly when r[pos] equals digit d.

The answer defines bestPos[v] as the maximum matched-prefix length reached so far at vertex v.

The answer includes a dominance pruning rule that discards a visit to vertex v when the new matched length is less than or equal to bestPos[v].

The answer includes an explicit acceptance condition that outputs YES when the matched length reaches L.

### 1) Build the functional graph of forced moves

Index each cell `(x,y)` as a vertex `v` (total `V = n*m ≤ 40000`).
Let `dig[v]` be its digit.

Compute the forced transition `to[v]`:
- Let `d = dig[v]`.
- Candidate `(nx, ny) = (x + a[d], y + b[d])`.
- If inside the grid: `to[v] = id(nx,ny)`, else `to[v] = v` (self-loop).

So the directed graph is **functional**: every vertex has outdegree exactly 1.
Hence each weakly connected component contains exactly one directed cycle, with directed trees feeding into it.

We will also build the reverse adjacency list:
\[
rev[u] = \{ v \mid to[v] = u \}.
\]

---

### 2) Convert the “write some visited digits” condition to a subsequence problem

Fix a starting vertex `start`. The forced movement generates an infinite walk:
\[
start, to[start], to[to[start]], \dots
\]
Wilbur can choose to write or skip at each visited vertex, so the produced string is **exactly a subsequence** of the digit sequence along this walk (and he may stop after finishing the string).

So for each query string `s`, we need to know whether there exists a start vertex whose walk contains `s` as a subsequence.

---

### 3) Reverse the query and run the process on the reversed graph

Let `s` be a query, length `L`, and let `r = reverse(s)`.

Instead of searching forward for `s` as a subsequence along `to[]`, we search for `r` as a subsequence along a walk in the **reversed graph** that starts on a cycle and moves outward:
- Reversing time turns “following `to[]`” into “choosing a predecessor from `rev[]`”.
- Cycles remain cycles after reversing edges.
- In the reversed functional graph, every component is a directed cycle with directed trees **going out of** the cycle.

Thus any reversed-time walk can:
1) loop around the cycle any number of steps (including 0),
2) then move outward along tree edges (cannot return to the cycle).

This matches the idea that forward-time walks eventually enter a cycle and then repeat it forever.

---

### 4) Precompute subsequence “next-state” table for the reversed string

For each query string `r` (length `L`), build:
- `nxt[pos][d]` for `pos = 0..L` and digit `d ∈ {0..9}`

Interpretation:
- `nxt[pos][d]` = the new position after greedily consuming digit `d` starting from `pos`:
  - if `pos < L` and `r[pos] == d`, then `nxt[pos][d] = pos+1`
  - else `nxt[pos][d] = pos`

(Equivalently, since we only need single-character consume/skip, this is just a constant-time transition.)

So consuming along any digit sequence becomes repeated application of `pos = nxt[pos][digit]`.

---

### 5) DFS with dominance pruning from cycles in the reversed graph

For a fixed query, we perform a reachability-style DP over states `(v, pos)`:
- `v` is a vertex,
- `pos` is how many characters of `r` have been matched so far.

We maintain an array `bestPos[v]` initialized to `-1`.
Meaning: we have already reached vertex `v` with some matched prefix length `bestPos[v]`, and any future arrival with `pos ≤ bestPos[v]` is useless (it cannot lead to matching more of `r` than what we already could from `v`).

DFS routine `dfs(v, pos)`:
1) If `pos == L`, we can stop: answer is `YES`.
2) Let `pos2 = nxt[pos][dig[v]]` (choose to write if it matches next, else effectively skip).
3) If `pos2 <= bestPos[v]`, return (dominated).
4) Set `bestPos[v] = pos2`.
5) For each predecessor `u` in `rev[v]`, call `dfs(u, pos2)`.

We must allow starting points that can loop on cycles to gain matches before leaving.
To handle looping, for each directed cycle we do:
- pick any cycle vertex as root,
- traverse the cycle edges in order and repeatedly relax `bestPos` on cycle vertices until no improvement occurs (at most `L` improvements total across the cycle because `bestPos` increases and is bounded by `L`),
- from the cycle vertices, launch `dfs` into outgoing tree edges (their `rev[]` children not on the cycle).

If any DFS reaches `pos == L`, print `YES`; otherwise `NO`.

Because `Σ = 10` and the graph size is at most `40000`, this pruning-based DFS runs fast in practice, and the total work is bounded by `O(V * Σ + L)` per query under the functional-graph structure and the monotonic `bestPos` pruning.

---

### 6) Output

Run the above per query string and print `YES` / `NO`.


Prompt:
You are given a tree with `n` vertices (`1` to `n`). Vertex `v` has color `c_v ∈ [1..n]`.

A simple path is **beautiful** if:
1. it has at least 2 vertices;
2. its endpoints have the same color;
3. no other vertex on the path has that endpoint color.

Paths are **undirected**.

### Input
- `n` (`2 ≤ n ≤ 2·10^5`)
- colors `c_1, ..., c_n`
- `n-1` edges

### Output
Output the number of beautiful simple paths.

Does the answer define a rooted tree (choose any root, e.g. 1) and process vertices in postorder (children before parent)?

Does the answer define cnt[v][x] as the number of vertices of color x in subtree(v) such that the path from v to that vertex contains no other vertex of color x (i.e., x-top-level w.r.t. v)?

Does the answer state that for any vertex v and any color x ≠ c[v], the number of beautiful paths whose LCA is v and whose endpoints have color x equals the sum over pairs of distinct children subtrees: (top-level x count in child A) × (top-level x count in child B)?

Does the answer state that for x = c[v], no beautiful path can have endpoints in two different child subtrees with LCA=v (because v would be an internal vertex of color x), so only paths (v, w) are allowed for x=c[v]?

Does the answer include the exact counting rule at vertex v for color c[v] as: add totalCv to the answer, where totalCv is the total number of c[v]-top-level vertices across all children of v?

Does the answer update the carried map at v by setting cnt[v][c[v]] = 1 (representing vertex v itself) and not propagating any descendant top-level vertices of color c[v] above v?

Does the answer describe a small-to-large (DSU-on-tree) merge where the largest child map is reused and all other child maps are merged into it, adding cross-pair contributions as ans += big[x] * small[x] for each x ≠ c[v] during the merge?

Does the answer state that each undirected beautiful path is counted exactly once by counting it at the LCA of its endpoints (or an equivalent unique-counting argument)?

Does the answer state a valid complexity bound for the DSU-on-tree implementation, e.g. O(n log^2 n) with ordered maps (or expected O(n log n) with hash maps), for n ≤ 2·10^5?

### Expert Solution

Root the tree arbitrarily (e.g. at vertex 1) and process vertices in postorder.

#### 1) Top-level counts
For each vertex `v`, maintain a map `cnt[v]` where:

`cnt[v][x] = #{ w ∈ subtree(v) : c_w = x and there is no other x on the path v → w }`

So `w` is **`x`-top-level w.r.t. `v`**.

This “top-level” property ensures that once a vertex of color `x` is encountered, deeper vertices of color `x` do not propagate above it.

#### 2) Unique counting by LCA
Count each undirected path exactly once at the **LCA** of its endpoints (in the chosen rooting).

Let `v` have children `u_1, ..., u_k`.

#### 3) Counting contributions at vertex v

- For any color `x ≠ c[v]`:
  choosing endpoints of color `x` from **two different child subtrees** creates a beautiful path with LCA `v`, because `v` is on the path but has color `≠ x`.
  The number of such paths is:
  `∑_{A<B} cnt[u_A][x] · cnt[u_B][x]`

- For color `x = c[v]`:
  endpoints **cannot** be in two different child subtrees with LCA `v`, since then `v` (color `x`) would be an internal vertex on the path, violating the rule.
  The only valid paths of color `c[v]` “through `v`” are `(v, w)` where `w` is `c[v]`-top-level in some child subtree. So add:
  `totalCv = ∑_{u child of v} cnt[u][c[v]]`
  to the answer.

#### 4) Small-to-large merge (DSU-on-tree)

To compute `cnt[v]` efficiently:

1. Pick the child `u*` with the largest map `cnt[u*]`. Let `big = cnt[u*]`.
2. Let
   `totalCv = big[c[v]]` (0 if absent)
   then erase key `c[v]` from `big` (descendant top-level `c[v]` must not propagate above `v`).
3. For each other child `u`:
   - For each `(x, val) ∈ cnt[u]`:
     - if `x = c[v]`: `totalCv += val` (do not merge)
     - else: add cross-pairs `ans += big[x] * val`, then `big[x] += val`
4. Add `ans += totalCv` (paths `(v, w)` with endpoint color `c[v]`).
5. Set `big[c[v]] = 1` because `v` itself becomes the only top-level occurrence of its color for ancestors.

Finally set `cnt[v] = big`.

#### 5) Complexity
Small-to-large guarantees each key is merged only `O(log n)` times.
- With ordered maps: `O(n log^2 n)`
- With hash maps (expected): `O(n log n)`

This fits `n ≤ 2·10^5`.


Prompt:
## Problem: Maximum Possible Score of One Player Under Greedy Picks

You are given `n` distinct items labeled `1..n`, with exactly one copy of each item. Each item `i` has an integer score `v[i]` (scores may be negative).

Two players, **A** and **B**, each have a complete ranking (a permutation) of the items:

- `A[1..n]`: player A’s preference list from most preferred to least preferred
- `B[1..n]`: player B’s preference list from most preferred to least preferred

The shop starts with all items available. Over exactly `n` turns, the following happens:

- On each turn, **either** A **or** B visits the shop (the order of visitors is not fixed).
- The visiting player takes the **highest-ranked item in their own list that is still available**.

After `n` turns, all items are taken and each player ends up with a set of items.

Your task: among all possible valid visit orders (i.e., all ways to choose who moves on each turn), compute the **maximum possible total score** of the items obtained by player **A**, where total score is the sum of `v[i]` over items A receives.

It can be proven that different visit orders can lead to different sets for A.

---

### Input

The first line contains an integer `t` — the number of test cases.
`1 ≤ t ≤ 10^4`

For each test case:

- An integer `n` (`1 ≤ n ≤ 2·10^5`)
- A line with `n` integers `v[1], v[2], ..., v[n]` (`-10^9 ≤ v[i] ≤ 10^9`)
- A line with permutation `A[1..n]` of `1..n`
- A line with permutation `B[1..n]` of `1..n`

Constraint: the sum of `n` over all test cases is at most `2·10^5`.

---

### Output

For each test case, output a single integer: the maximum possible sum of scores of items that player A can end up with.

Does the response define p[i] as Bob’s position of item a[i] in Bob’s permutation?

Does the response state the DP initialization dp[0] = 0?

Does the response state that the final answer equals max(dp[j]) over j from 0 to n?

Does the response specify a segment-tree range-maximum query over the index interval j in [0, p[i]−1] to compute bestPrefix for item i?

Does the response define a relabeling where item i corresponds to the original item a[i]?

Does the response define a DP state dp[j] as the maximum total value Alice can obtain with current maximum Bob-rank equal to j?

Does the response state that assigning item i to Alice is allowed only for DP states with j < p[i]?

Does the response state that assigning item i to Bob updates the maximum Bob-rank to max(j, p[i])?

Does the response specify a segment-tree range-add update of +val[i] applied to dp[j] over the index interval j in [0, p[i]−1]?

Does the response specify the point update dp[p[i]] := max(dp[p[i]], bestPrefix) after processing item i?

### 1) Renaming so that Alice’s list becomes `1,2,...,n`

Let the original item IDs be `1..n`. Define:

- `a[1..n]`: Alice’s permutation (most preferred first)
- `b[1..n]`: Bob’s permutation
- `v[id]`: value of item `id`

Relabel items by **Alice-rank**:

- The item with Alice-rank `i` is `id = a[i]`.
- Define `val[i] = v[a[i]]`.
- Let `pos[id]` be Bob’s rank of item `id` in `b` (i.e., `pos[b[j]] = j`).
- Define `p[i] = pos[a[i]]`.

After this transformation, we can think of items as `1..n` in Alice’s priority order, with:
- value `val[i]`
- Bob-rank `p[i]`

This does not change which outcomes are possible; it only renames items.

---

### 2) Which subsets can Alice end up with?

Suppose Alice finally gets a set `S` of Alice-ranks. Consider any pair:
- `x ∉ S` (Bob gets `x`)
- `y ∈ S` (Alice gets `y`)
with `x < y` (Alice likes `x` more than `y`).

A necessary condition is:

> `p[x] < p[y]`.

Reason:
If `p[x] > p[y]`, then Bob prefers `y` earlier than `x`, but Alice prefers `x` earlier than `y`. Under greedy picking, neither `x` nor `y` can be taken first without blocking the other, so such a final split is impossible.

It turns out this condition is also sufficient: if it holds for all such pairs, a valid visit order exists that realizes exactly `S` (you can schedule Bob to “clear” items until Alice’s next desired item becomes her current greedy choice).

So the problem becomes:

> Choose a subset `S` maximizing `sum(val[i] for i in S)` such that
> for every `x ∉ S`, `y ∈ S`, `x < y` implies `p[x] < p[y]`.

---

### 3) DP formulation using the maximum Bob-rank seen so far

Process items in increasing Alice-rank `i = 1..n`.

Let `dp[j]` mean:

- after deciding ownership of the first `i` items,
- the maximum Bob-rank among items assigned to Bob equals `j`,
- `dp[j]` is the maximum achievable sum of values assigned to Alice.

Here `j` ranges from `0..n`, where `j = 0` means Bob has taken nothing yet.

Initialization:
- `dp[0] = 0`
- `dp[j] = -∞` for `j > 0`

Now process item `i` with Bob-rank `p = p[i]` and value `val = val[i]`.

**Case A: Alice takes item `i`**
This is allowed exactly when `p > j` (all earlier Bob-taken items have smaller Bob-rank).
State `j` stays the same, value increases:

- for all `j < p`: `dp_next[j] = dp[j] + val`

**Case B: Bob takes item `i`**
Value does not change, new max Bob-rank becomes `max(j, p)`:

- if `j ≥ p`: it remains in state `j`
- if `j < p`: it moves to state `p`

---

### 4) Segment tree optimization (O(n log n))

The transition above can be implemented in-place on a segment tree over indices `j = 0..n` storing current `dp[j]`.

For item `(p, val)`:

1. Compute:
   - `bestPrefix = max(dp[j])` over `j in [0, p-1]`

2. Apply Alice-taking updates:
   - For all `j in [0, p-1]`, do `dp[j] += val`
   - This is a **range add**.

3. Apply Bob-taking update into state `p`:
   - `dp[p] = max(dp[p], bestPrefix)`
   - This is a **point chmax** after a range max query.

Note:
- States `j > p` are unchanged in this step.
- State `p` already contains the `j = p` “Bob keeps max = p” option, and the `bestPrefix` handles all `j < p` moving into `p`.

After processing all `n` items, the answer is:
- `max(dp[j])` over `j = 0..n`

Data structure requirements:
- range add
- range maximum query
- point update to set `dp[p] := max(dp[p], x)`

A lazy segment tree supports these in `O(log n)` each, giving total complexity `O(n log n)` per test file (with `sum n ≤ 2·10^5`).

---

### 5) Summary of implementation steps

For each test case:
1. Read `v`, `a`, `b`.
2. Build `pos[id]` from `b`.
3. Build arrays for `i = 1..n`:
   - `val[i] = v[a[i]]`
   - `p[i] = pos[a[i]]`
4. Initialize segment tree with `dp[0]=0`, others `-∞`.
5. For `i=1..n`, perform the three operations described above.
6. Output the maximum stored `dp[j]`.


Prompt:
You have a strip of `1 × 10^9` cells numbered `1` to `10^9`.

There are `n` snakes (numbered `1` to `n`). Initially, each snake occupies exactly one cell, and no two snakes can be placed in the same cell. After placement, the game runs for `q` seconds. Each second exactly one event occurs:

- **Snake `s_i` enlarges:** if it currently occupies `[l, r]`, it becomes `[l, r+1]`.
- **Snake `s_i` shrinks:** if it currently occupies `[l, r]`, it becomes `[l+1, r]`.

If at any time any snake hits an obstacle (another snake or the end of the strip), you lose. Otherwise, you win with a score equal to the maximum cell index occupied by any snake at any time.

Given the sequence of `q` events (guaranteed valid: a snake of length 1 never shrinks), find the minimum possible score over all initial placements.

## Input
- First line: integers `n, q` (`1 ≤ n ≤ 20`, `1 ≤ q ≤ 2 · 10^5`).
- Next `q` lines: either `"s_i +"` or `"s_i -"`.

## Output
Output one integer — the minimum possible score.

Does the solution state the final answer as min over last of ( d[(1<<n)-1][last] + plusCount[last] ), where plusCount[k] is the number of '+' events of snake k?

Does the solution state the score for a fixed permutation p1..pn as 1 + (sum of gaps between consecutive snakes) + (number of '+' events of the last snake pn)?

Does the solution state that choosing an order of snakes on the strip corresponds to choosing a permutation of the n snakes?

Does the solution define (for an ordered pair (i,j) with j immediately to the right of i) that an event contributes e=−1 when snake i enlarges and e=+1 when snake j shrinks (and e=0 otherwise) to the evolution of the gap?

Does the solution state the DP base case d[1<<i][i] = 1 for every snake i?

Does the solution state that for a fixed permutation, it is optimal to place snakes greedily as close together as possible (i.e., using the minimum feasible gaps between consecutive snakes)?

Does the solution give the exact formula for minDist[i][j] as 1 − min_prefix_sum of the cumulative e-values over all events (equivalently, minDist[i][j] = -min_{1..q})?

Does the solution define the DP state d[mask][last] as the minimum possible starting cell (position) of snake 'last' after placing exactly the set 'mask' in some order, with the first placed snake starting at cell 1?

## Expert Answer

Key idea: placing snakes left-to-right defines a permutation; for any fixed order the best placement is greedy (minimum feasible gaps), and we can compute the required minimum gaps from the event log.

### 1) Reduce to an order (permutation)
In any valid placement, the snakes appear left-to-right in some permutation `p_1, p_2, ..., p_n`. If this permutation is fixed, it is optimal to place them as close as possible (use the smallest initial gaps that still avoid collisions during all events), because any extra initial spacing only increases the maximum occupied cell.

### 2) Score for a fixed permutation
Let the initial gap between consecutive snakes `p_k` and `p_{k+1}` be `g_k`. Then:
`score(p) = 1 + ∑_{k=1..n-1} g_k + plusCount(p_n)`
because the first snake can start at cell 1, the last snake starts at `1 + ∑ g_k`, and only enlargements of the last snake can push the global maximum further right.

### 3) Compute `minDist[i][j]` for "`j` immediately right of `i`"
Assume snake `j` is immediately to the right of snake `i` with initial gap `x`. The gap changes only when:
- snake `i` enlarges (gap `-1`),
- snake `j` shrinks (gap `+1`).

Define per-event contribution `e_t` for ordered pair `(i, j)`:
- `e_t = -1` if event `t` is "`i` +",
- `e_t = +1` if event `t` is "`j` -",
- `e_t = 0` otherwise.

After `t` events the gap is `x + ∑_{u=1..t} e_u`. We need this `≥ 0` for all `t`, so:
`minDist[i][j] = 1 - min_{1 ≤ t ≤ q} (∑_{u=1..t} e_u)`
(This equals `-min(...) + 1`).

### 4) Choose the best permutation via bitmask DP
Let `d[mask][last]` be the minimum possible starting cell of snake `last` after placing exactly the set `mask` in some order, with the first placed snake starting at cell 1.

Base:
`d[1 << i][i] = 1` for all `i`.

Transition:
`d[mask | (1 << j)][j] = min_{i ∈ mask} (d[mask][i] + minDist[i][j])`

Final answer:
`min_{1 ≤ last ≤ n} ( d[(1 << n) - 1][last] + plusCount(last) )`


Prompt:
## Problem

Denis defines a non-commutative “multiplication” of strings.
For a string `s` of length `m` and a string `t`, define:

\[
s \cdot t = t + s_1 + t + s_2 + \dots + t + s_m + t
\]

where \(s_i\) is the \(i\)-th character of `s`, and `+` means concatenation.

Examples:
- `"abc" ⋅ "de" = "deadebdecde"`
- `"ab" ⋅ "z" = "zazbz"`

Roman defines the **beauty** of a string as the length of the longest substring consisting of only one repeated character (i.e., the maximum run length).

You are given `n` strings \(p_1, p_2, \dots, p_n\). Compute the beauty of:

\[
(\dots(((p_1 \cdot p_2)\cdot p_3)\cdot \dots)\cdot p_n)
\]

It is guaranteed that the beauty of the final product is at most \(10^9\).

### Input
- First line: integer \(n\) \((2 \le n \le 10^5)\)
- Next \(n\) lines: non-empty strings \(p_1,\dots,p_n\) of lowercase English letters

Constraints:
- Total length of all \(p_i\) is at most \(10^5\)
- Final beauty is at most \(10^9\)

### Output
Print one integer — the beauty of the final product.

The answer states that the operation satisfies associativity: (a·b)·c = a·(b·c).

The answer computes the product by folding from right to left using P_n = p_n and P_i = p_i · P_{i+1}?

The answer introduces a capping constant INF equal to 10^9 (or an explicitly stated value greater than or equal to 10^9) for tracked lengths.

The answer defines a boolean field allSame(X) that indicates whether all characters in X are equal.

The answer defines the concatenation cross-run value as A.suffLen + B.prefLen when A.suffChar equals B.prefChar.

The answer defines best(A+B) as the maximum of A.best, B.best, and the cross-run value for the concatenation boundary.

The answer gives the prefix-length update rule prefLen(A+B) = A.len + B.prefLen when A.allSame is true and A.prefChar equals B.prefChar.

The answer gives the suffix-length update rule suffLen(A+B) = B.len + A.suffLen when B.allSame is true and A.suffChar equals B.suffChar.

The answer describes computing s·t by initializing res as the summary of t and then applying the loop res=merge(res,single(c)) followed by res=merge(res,tSummary) for each character c of s.

The solution states an overall runtime of O(total input length) (i.e., O(∑|pi|)) for the full computation.

### 1) Use associativity and fold from right to left

The operation is associative:
\[
(a\cdot b)\cdot c = a\cdot(b\cdot c)
\]
(you can verify by expanding both sides; both produce alternating blocks of the right operand with single characters of the left operands inserted in the same order).

So instead of building the product left-to-right (which can temporarily explode and later “collapse”), we compute:
\[
P_i = p_i \cdot P_{i+1}, \quad P_n = p_n
\]
from \(i=n-1\) down to \(1\).

Because the final beauty is guaranteed \(\le 10^9\), when folding right-to-left we can safely cap all tracked lengths at \(INF=10^9\) (or \(10^9+5\)) to avoid overflow without affecting the answer.

---

### 2) Maintain only a compact summary of a string

For any (possibly huge) string \(X\), store the tuple:

- `len(X)` (capped to `INF`)
- `best(X)`: maximum run length anywhere in \(X\)
- `prefChar(X)`, `prefLen(X)`: character and length of the maximal constant prefix
- `suffChar(X)`, `suffLen(X)`: character and length of the maximal constant suffix
- `allSame(X)`: whether all characters in \(X\) are equal

This is sufficient to compute the same tuple for concatenations.

For a single character `c`:
- `len=1`, `best=1`
- `prefChar=suffChar=c`, `prefLen=suffLen=1`
- `allSame=true`

For an explicit input string, compute the tuple by a linear scan.

---

### 3) Concatenation merge routine

Let `merge(A,B)` return the summary for the concatenation \(AB\). With `INF` capping:

- `len = min(INF, A.len + B.len)`
- `prefChar = A.prefChar`
- `prefLen`:
  - if `A.allSame` and `A.prefChar == B.prefChar` then `min(INF, A.len + B.prefLen)`
  - else `A.prefLen`
- `suffChar = B.suffChar`
- `suffLen`:
  - if `B.allSame` and `A.suffChar == B.suffChar` then `min(INF, B.len + A.suffLen)`
  - else `B.suffLen`
- `allSame = A.allSame && B.allSame && (A.prefChar == B.prefChar)`
- `best = max(A.best, B.best, cross)` where
  - `cross = (A.suffChar == B.prefChar) ? min(INF, A.suffLen + B.prefLen) : 0`

---

### 4) How to compute `s ⋅ t` using only merges

By definition:
\[
s\cdot t = t + s_1 + t + s_2 + \dots + t + s_m + t
\]

So if we already have the summary of \(t\) as `T`, and `s` is an explicit string of length `m`, we can build the summary of \(s\cdot t\) as:

- `res = T`
- for each character `c` in `s` from left to right:
  - `res = merge(res, single(c))`
  - `res = merge(res, T)`

This performs \(O(|s|)\) merges.

---

### 5) Full algorithm

- Read all strings \(p_1,\dots,p_n\).
- Set `cur = summary(p_n)`.
- For `i = n-1` down to `1`:
  - Let `T = cur` (summary of the current right operand).
  - Compute `cur = summary(p_i ⋅ current)` by the loop in section 4 using merges and `T`.
- Output `cur.best`.

Complexity:
- Each character of each input string is processed a constant number of times.
- Total time \(O(\sum |p_i|)\), total memory \(O(1)\) besides storing the input and small summaries.


Prompt:
An autonomous vehicle navigation system uses a bidirectional time-dependent Dijkstra algorithm to compute an optimal route on a directed graph
G with 12 nodes labeled N0 through N11 and the following directed edges with base travel times in seconds:

(N0,N1,45), (N0,N2,60), (N1,N3,50), (N1,N4,70), (N2,N4,55), (N2,N5,80),
(N3,N6,65), (N4,N6,40), (N4,N7,90), (N5,N7,60), (N5,N8,75),
(N6,N9,55), (N7,N9,50), (N7,N10,85), (N8,N10,70),
(N9,N11,45), (N10,N11,40).

Travel times depend on the start time of traversal of each edge as follows:

• For any traversal that begins between t = 0 and t = 120 seconds, if the edge is incident to node N4 (either entering or leaving N4), its base weight is multiplied by 1.8.
• For any traversal that begins between t = 120 and t = 240 seconds, if the edge is incident to node N7, its base weight is multiplied by 2.2.
• Outside these windows, all edges use their base weights.

The vehicle departs from node N0 at time t = 0.

Algorithm Definition (Deterministic)

The algorithm runs a forward search from N0 and a backward search from N11 using the following rules:

1. Priority Queues

Each search maintains its own min-heap priority queue ordered by cumulative cost.
Ties inside a queue are broken by smaller node index (e.g., N3 < N4).

2. Expansion Scheduling

At each iteration, the algorithm compares the minimum keys in the forward and backward priority queues and expands the node with the smaller key.
If the two minimum keys are equal, the forward queue is expanded.

3. Forward Search Cost Model

The forward search computes exact arrival times using the time-dependent cost rules above.
When relaxing an edge (u,v), the traversal start time equals the arrival time at u, and the edge cost is multiplied if u or v is N4 or N7 and the traversal start time lies in the corresponding time window.

4. Backward Search Cost Model (Before First Complete Path)

Before any complete N0→N11 path is discovered, the backward search computes distances using base edge weights only, treating them as admissible lower bounds.

5. Backward Search Cost Model (After First Complete Path)

After a complete path with total cost C* is discovered, the backward search recomputes all subsequent relaxations using the same time-dependent cost model as the forward search, assuming arrival at N11 occurs at time C*.

6. Meeting Condition

A node is defined to be a meeting node when it has been expanded (popped from the priority queue) in both the forward and backward searches.

The first such node encountered is recorded as the meeting node.

7. Path Reconstruction

The full path is constructed by concatenating the forward parent chain from N0 to the meeting node and the backward parent chain from the meeting node to N11.

8. Termination Condition

The algorithm terminates when:

min_key_forward + min_key_backward > best_complete_path_cost

where:
• min_key_forward is the smallest key currently in the forward queue,
• min_key_backward is the smallest key currently in the backward queue,
• best_complete_path_cost is the lowest-cost N0→N11 path found so far.

These values are evaluated immediately after each node expansion.

9. Priority Queue Insertions

Every insertion into either priority queue is counted, including duplicate insertions caused by cost improvements.

Task

Determine:

(1) the exact sequence of nodes expanded by the forward search until termination,
(2) the exact sequence of nodes expanded by the backward search until termination,
(3) the first meeting node,
(4) the optimal path from N0 to N11,
(5) the total travel time of that path in seconds, and
(6) the total number of nodes inserted into priority queues across both searches.

Does the answer state the forward expansion sequence exactly as [N0, N1, N2, N3, N5, N4, N6, N8, N9]?

Does the answer state the backward expansion sequence exactly as [N11, N10, N9, N7, N6, N8, N5, N4]?

Does the answer identify N9 as the first meeting node?

Does the answer give the optimal path exactly as N0 → N1 → N3 → N6 → N9 → N11?

Does the answer compute the total travel time as exactly 260 seconds?

Does the answer report the total priority queue insertion count as exactly 24?

Does the answer explicitly show that no N4- or N7-incident edges are used in the optimal path?

Does the answer demonstrate that termination occurs because min_forward_key + min_backward_key > 260?

(1) Forward Expansion Sequence

Forward search starts at N0 at time t = 0 and applies time-dependent weights.

Expand N0 (0)
→ insert N1(45), N2(60)

Expand N1 (45)
→ insert N3(95), N4(171) [70 × 1.8 due to N4 window]

Expand N2 (60)
→ update N4(159) [55 × 1.8], insert N5(140)

Expand N3 (95)
→ insert N6(160)

Expand N5 (140)
→ insert N7(272) [60 × 2.2 due to N7 window], insert N8(215)

Expand N4 (159)
→ attempt update N6(199), ignored (existing 160 is better)

Expand N6 (160)
→ insert N9(215)

Expand N8 (215)
→ insert N10(285)

Expand N9 (215)
→ insert N11(260)

At this point, best complete path cost = 260.
Next forward key = 272, next backward key = 155, and
272 + 155 > 260 ⇒ termination.

Forward Expansion Order:
[N0, N1, N2, N3, N5, N4, N6, N8, N9]

(2) Backward Expansion Sequence

Backward search begins at N11 using base weights until first complete path.

Expand N11 (0)
→ insert N9(45), N10(40)

Expand N10 (40)
→ insert N7(125), N8(110)

Expand N9 (45)
→ update N7(95), insert N6(100)

Expand N7 (95)
→ insert N4(185), insert N5(155)

Expand N6 (100)
→ insert N3(165)

Expand N8 (110)
→ no improvements

Expand N5 (155)
→ insert N2(235)

Expand N4 (185)
→ insert N1(255)

Termination condition satisfied:
272 + 155 > 260.

Backward Expansion Order:
[N11, N10, N9, N7, N6, N8, N5, N4]

(3) First Meeting Node

N9 is expanded in:
• forward search after N6
• backward search after N11

No earlier node is expanded in both searches.

Meeting Node: N9

(4) Optimal Path

Forward parents:
N0 → N1 → N3 → N6 → N9

Backward parents:
N9 → N11

Combined path:

N0 → N1 → N3 → N6 → N9 → N11

(5) Total Travel Time

No N4 or N7 edges appear on this path, so no multipliers apply.

N0 → N1 = 45
N1 → N3 = 50
N3 → N6 = 65
N6 → N9 = 55
N9 → N11 = 45

Total:

45 + 50 + 65 + 55 + 45 = 260 seconds

(6) Total Priority Queue Insertions

Each discovery or cost-improving update counts.

Forward insertions (12):
N1, N2
N3, N4
update N4, N5
N6
N7, N8
N9
N10
N11

Backward insertions (12):
N9, N10
N7, N8
update N7, N6
N4, N5
N3
N2
N1

Total insertions:

12 + 12 = 24


Prompt:
A miner is searching a rooted underground shaft system that forms a **tree** of `n` chambers (nodes) connected by one-way-less tunnels (undirected edges). A single phantom hides in exactly one chamber; initially it may be in any chamber. The miner can perform two kinds of actions. After each action, the phantom may (or may not) move to one adjacent chamber, except it is forbidden to move into the chamber that the miner just inspected in that same action.

Available actions (each targets a single chamber `x`):

- `1 x` — **Inspect** chamber `x`. If the phantom is currently in `x`, the miner catches it and the process ends. Otherwise the phantom may move to any adjacent chamber except `x`.
- `2 x` — **Seal** chamber `x`: destroy every tunnel incident to `x`. After sealing, the phantom cannot use those destroyed tunnels. Then the phantom may move to one adjacent chamber (along any remaining tunnels).

Design a single strategy — i.e., an explicit sequence of actions — that guarantees the miner will catch the phantom regardless of its initial position and its choices of moves. Your sequence must contain at most `⌊5n/4⌋` actions.

## Input

The input contains multiple test cases. The first line contains a single integer `t` (`1 ≤ t ≤ 10^4`) — the number of test cases. The sum of `n` over all test cases does not exceed `2·10^5`.

Each test case has the following format:

- A line with an integer `n` (`2 ≤ n ≤ 2·10^5`) — the number of chambers.
- Then `n-1` lines follow, each with two integers `u` and `v` (`1 ≤ u, v ≤ n`), indicating an undirected tunnel between chambers `u` and `v`. The tunnels form a tree.

## Output

For each test case, output a sequence of actions that will surely catch the phantom.

First print a single integer `k` (`1 ≤ k ≤ ⌊5n/4⌋`) — the number of actions in your strategy.

Then print `k` lines; the `i`-th such line should contain two integers `ti` and `xi`:

- `ti` is either `1` or `2`, specifying the action type (inspect or seal).
- `xi` is the target chamber index (`1 ≤ xi ≤ n`).

If the miner inspects a chamber and the phantom happens to be there, the process stops immediately (no further actions are performed). Your printed sequence should therefore be a valid plan that would succeed in every possible adversarial scenario (the grader will verify your action list satisfies the bound and would lead to capture irrespective of phantom moves).

## Constraints summary

- `1 ≤ t ≤ 10^4`
- `2 ≤ n ≤ 2·10^5`
- Sum of `n` across all test cases ≤ `2·10^5`
- Output sequence length per test case must satisfy `k ≤ ⌊5n/4⌋`

## Notes

- You do **not** need to minimize `k`, only ensure `k ≤ ⌊5n/4⌋` and that the strategy guarantees capture.
- The judge will simulate the worst-case adversarial phantom behavior to check your plan.

Does the solution include a postorder coloring procedure that specifies how to label each node as 'green', 'yellow', or 'black' using only information from its children?

Does the solution assert that after removing every node labeled 'black', every connected component of the remaining graph is a simple path (every node has residual degree ≤ 2)?

Does the solution state that groups associated with different black nodes are pairwise disjoint?

Does the solution derive the inequality n ≥ 4·b where b denotes the number of black nodes?

Does the solution state that when a node is colored black because it has three or more green children, it is grouped with three distinct green-subtree endpoint vertices (i.e., the solution explicitly assigns three distinct green vertices to that black node)?

Does the solution state that the worst-case total number of operations equals n + b and then conclude this value is ≤ ⌊5n/4⌋?

Does the solution state the algorithmic resource bounds as O(n) time and O(n) memory for producing the operation sequence?

Does the solution explicitly state that all type-2 (cut) operations are performed before any type-1 (inspect) operations?

Does the solution describe the sweep procedure for a path as performing exactly one type-1 operation per node while checking nodes in sequence from an endpoint to the other endpoint?

Does the solution explicitly justify that moving all type-2 (seal) operations earlier cannot reduce correctness because sealing only restricts the phantom’s movement options?

**Idea in one line.**
Cut a carefully chosen small set of nodes so the remaining connected pieces are simple paths, then “sweep” each path by checking its nodes from one end to the other. Do all cuts first (they only restrict the phantom), then do the sweeps. The number of cuts is at most `⌊n/4⌋`, so the total number of actions (cuts + checks) is `≤ n + ⌊n/4⌋ = ⌊5n/4⌋`.

---

## Full algorithm (constructive)

1. **Root the tree** at node `1`. We will do a single postorder DFS to compute one of three colors for every node: **green**, **yellow**, or **black**.

2. **Coloring rule (postorder):** for each node `v`, let `G` = number of children of `v` colored **green**, and let `Y` = whether `v` has any child colored **yellow**.
   - If `G ≥ 3` **or** `Y = true`, color `v` **black**.
   - Else if `G = 2` (and `Y = false`), color `v` **yellow**.
   - Else (i.e., `G ≤ 1` and `Y = false`), color `v` **green**.

   Compute these colors by visiting children first (postorder), so each child’s color is known when we decide the parent’s color.

3. **Perform all cuts.** For every node `v` colored **black**, output operation `2 v` (destroy all edges incident to `v`). Do these `2`-operations first (in any order).

   After cutting every black node, every connected component of the remaining graph is a simple path (this follows from how colors were assigned — nodes with degree ≥3 in the leftover graph were prevented by the coloring).

4. **Sweep each path.** For each connected component (which is now a path of length `k` nodes), choose one endpoint and perform `1 x` operations along the path from that endpoint to the other endpoint (i.e., check first node, then next, …, last), listing `k` operations for that path. Do these sweeps after finishing all `2`-operations.

   If any `1 x` finds the phantom, the process stops early (the plan is still valid since we listed a complete schedule for the adversary-free guarantee).

---

## Why this works

- **Doing all `2` first is safe.** Cutting edges only reduces the phantom’s possible moves. Sealing a node only deletes tunnels, so it can only reduce the phantom’s set of legal moves and never creates new moves; therefore performing type-2 operations earlier cannot make capture harder. Hence we may assume all `2`-operations come before any `1`-operation.

- **After the cuts every component is a path.** The coloring rule forbids any node (in the uncut graph) from having three or more “green” child-subpaths survive in its subtree without itself being black. When we cut out every black node (remove all incident edges), every remaining component has no branching — it must be a path (degree ≤ 2 everywhere). Informally: yellow nodes serve only to join two green child-paths into one longer path; black nodes break branches.

- **Sweeping a path of length `k` with `k` checks works.** If you check nodes along the path from one endpoint to the other using `1`-operations, after `i` checks the phantom cannot be located at any of the first `i` nodes (because if it were, it would be caught at that check; otherwise it could not cross back past the last checked node because edges behind black nodes are removed). By induction this eventually forces capture.

---

## Operation count (bound)

- Let `b` be the number of black nodes produced by the coloring. We perform exactly `b` operations of type `2` (one for each black node).

- After removing edges incident to black nodes, the graph splits into disjoint paths covering the `n` nodes. Sweeping every path requires exactly `n` operations of type `1` in total (each node is checked exactly once in the worst-case adversarial schedule). Thus worst-case total operations = `n + b`.

- **Key combinatorial bound:** The coloring guarantees `b ≤ ⌊n/4⌋`. Sketch of charging argument:
  - Every time a node becomes black because it had `≥ 3` green children, choose exactly three of those green children and charge one distinct green-subtree endpoint from each of the three to that black node.
  - Every time a node becomes black because it had a yellow child, charge the yellow child and its (up to) two green children — again three distinct nodes charged to this black node.
  - The groups associated with different black nodes are pairwise disjoint (no vertex is charged to two different black nodes).
  - Hence each black node accounts for at least 4 distinct vertices (the black node itself + at least 3 charged vertices), so n ≥ 4·b.

- Therefore total operations `n + b ≤ n + ⌊n/4⌋ = ⌊5n/4⌋`, meeting the required bound.

---

## Complexity and implementation notes

- The coloring (postorder DFS) runs in `O(n)` time and uses `O(n)` memory (recursion or an explicit stack).
- Producing the list of cut operations is `O(b)`. After removing black nodes conceptually, extracting the path components and listing sweep orders is `O(n)`.
- Overall time complexity `O(n)` per test case; total across all tests `O(sum n)`.

---

## Summary (what to output)

- First, output `k = n + b` operations (or fewer — the schedule may stop early if the phantom is caught).
- Then print:
  1. `2 v` for every black node `v` (order arbitrary).
  2. For each remaining path component, a sequence `1 x1`, `1 x2`, …, `1 xk` checking nodes along the path from one endpoint to the other endpoint.

This produces a valid plan that always catches the phantom and uses at most `⌊5n/4⌋` operations.


Prompt:
An Arbor Network is an undirected, connected, acyclic graph (i.e., a tree) with N junctions numbered 1 through N.

Every morning, R rival couriers traverse the network simultaneously. Rival i starts at junction s_i and travels to junction t_i along the unique shortest path in the tree. If that path is the vertex sequence p_1, p_2, ..., p_k with p_1 = s_i and p_k = t_i, then at integer time j (for 1 ≤ j ≤ k) the rival is exactly at vertex p_j.

After a rival reaches t_i, they stop moving and cannot be encountered thereafter. Concretely, for rival i, encounters are only possible at times 2, 3, ..., k (while the rival is still traveling); times after reaching the destination are ignored for collision purposes.

A protagonist courier starts at junction A at time 1 and must reach junction B (A ≠ B). At each integer time step, the protagonist may either:

move along one incident road to an adjacent junction, or
wait at the current junction.
A collision occurs if at any integer time t the protagonist and any rival occupy the same junction. (Being on the same edge segment at the same time is allowed.)

It is guaranteed that s_i ≠ A for all rivals i.

Goal:
For each test instance, compute the earliest integer time at which the protagonist can arrive at junction B without ever colliding at a junction with any rival courier. If it is impossible, output -1.

Input:
The first line contains an integer T, the number of test instances.

Each test instance:

One line with four integers N, R, A, B (2 ≤ N ≤ 10^5, 1 ≤ R ≤ 200, 1 ≤ A,B ≤ N, A ≠ B)
Next N-1 lines: each line has two integers u and v describing an undirected edge of the tree
Next R lines: each line has two integers s_i and t_i with s_i ≠ t_i and s_i ≠ A
It is guaranteed that the sum of all N across test instances does not exceed 10^5.

Output:
For each test instance print one integer: the earliest collision-free arrival time at B, or -1.

Does the answer explicitly state the finite search horizon bound t* ≤ 2N + 1?

Does the answer explicitly define a forbidden pair (u, t) as: some rival occupies vertex u at integer time t during its travel?

Does the answer explicitly state the forbidden-pair budget: the total number of forbidden (vertex, time) pairs is at most N·R?

Does the answer explicitly define the base frontier/DP set as S1 = {A}?

Does the answer explicitly state the waiting transition: if v ∈ S_t and (v, t+1) is not forbidden, then v ∈ S_{t+1}?

Does the answer explicitly state the movement transition: if u ∈ S_t, v is adjacent to u, and (v, t+1) is not forbidden, then v ∈ S_{t+1}?

Does the answer explicitly state that the reported arrival time equals min{ t : B ∈ S_t}?

Does the answer explicitly state the failure rule: output −1 if there is no t ≤ 2N + 1 such that B ∈ S_t?

Does the answer explicitly state the combinatorial bound that each vertex can enter the frontier S_t at most R + 1 times over all t?

Does the answer explicitly state the overall time complexity bound O(N·R)?

Explicit finite time horizon: t* ≤ 2N + 1
Let t* be the earliest arrival time at B among all collision-free schedules, if any exist.

Claim. If a collision-free schedule exists, then t* ≤ 2N + 1.

Reason (sketch). Each rival walks a simple tree path of length at most N, so all rivals have finished moving by time ≤ N, hence encounter constraints cease no later than time N + 1. From time N + 1 onward, the protagonist can ignore rivals and reach any node (in particular B) within at most N more moves in a tree. Therefore some collision-free schedule (if one exists at all) can be made to arrive by (N + 1) + N = 2N + 1. So we only need to search times t ≤ 2N + 1.

Forbidden pairs and the budget ≤ N · R
Define a vertex-time pair (u, t) as forbidden if at time t some rival occupies vertex u at a time when encounters count (i.e., along their travel times 2 through the path length).

Each rival visits at most N vertices over time, hence contributes at most N forbidden pairs. Therefore the total number of forbidden pairs satisfies:
{(u, t) forbidden} ≤ N · R.

Store forbidden information as per-time lists: forbidden[t] = list of vertices forbidden at time t. Total stored entries is O(NR).

Formal DP / frontier recurrence (explicit)
Let S_t ⊆ {1, ..., N} be the set of vertices where the protagonist can be at time t without having collided at any time ≤ t.

Base:
S_1 = {A}.

Transition (for every t ≥ 1): for each vertex v,
v ∈ S_{t+1} ⇔ ((v, t+1) is not forbidden) AND (v ∈ S_t OR ∃u ∈ N(v) with u ∈ S_t).
This exactly models “wait or move one edge” while avoiding forbidden vertex-time collisions.

Output rule (explicit):
answer = min { t : B ∈ S_t },
and output -1 if no such t exists with t ≤ 2N + 1.

Key invariant: each vertex enters at most R + 1 times
Use an incremental frontier computation (do not build a full N × (2N) DP table).

Invariant (explicit). Over all times, any fixed vertex v can become newly feasible (i.e., “enter” the frontier S_t) at most R + 1 times.

Justification (sketch): a vertex can be removed from feasibility at time t only if it is forbidden at t. A given vertex can be forbidden by at most R rivals (each at some time), hence it can be “kicked out” at most R times, so it can enter at most R + 1 times.

Incremental processing and total work bounds
Process time steps t = 1, 2, ..., 2N + 1 while maintaining the current feasible set S_t (as a boolean in[v]) and a candidate list for membership tests.

Type-1 removals handled incrementally (explicit). At time t, mark all vertices in forbidden[t] as blocked for time t. The total amount of such removal handling over the whole run is:
∑_t |forbidden[t]| ≤ N · R.

Neighbor-announcement bound (explicit). Each time a vertex v enters the frontier, we may “announce” its neighbors as candidates for the next time. Since v enters at most R + 1 times, the total neighbor-announcement work is bounded by:
∑_v deg(v) · (R + 1) = (R + 1) ∑_v deg(v) = (R + 1) · 2(N − 1) ≤ 2N(R + 1).

Total candidate processing (explicit). Candidates arise only from:

forbidden-event triggers (bounded by N · R), and
neighbor announcements (bounded by 2N(R + 1)).
Thus, the total number of candidate checks processed over all times is O(NR).
Final complexities (explicit)
Time complexity: O(NR), justified by combining:

forbidden-event processing ≤ N · R, and
neighbor announcements ≤ 2N(R + 1),
plus O(1)-amortized candidate checks per announced item.
Memory complexity: O(NR) for storing forbidden per-time lists (total entries ≤ N · R) plus O(N) for the tree adjacency and frontier arrays; overall O(NR).


Prompt:
Consider an n x m grid (2 <= n <= 12, 2 <= m <= 1000) where some cells are blocked. A Hamiltonian path visits every non-blocked cell exactly once. The task is to count Hamiltonian paths from cell (0,0) to cell (n-1, m-1), modulo 10^9+7. The grid is given as n lines of m characters where '.' is empty and '#' is blocked. If (0,0) or (n-1, m-1) is blocked, the answer is 0.

Describe a complete algorithm to solve this problem using broken profile dynamic programming (plug DP). Your answer must address ALL of the following points using the conventions specified below.

CONVENTIONS:
- The grid has n rows and m columns, 0-indexed.
- Cells are processed in column-major order (column 0 top-to-bottom, then column 1 top-to-bottom, etc.).
- The plug array is 0-indexed with exactly (n+1) entries.
- Component labels in the plug array are canonicalized by assigning integers 1, 2, 3, ... in left-to-right first-occurrence order.

ADDRESS EACH OF THE FOLLOWING:

1. PROFILE STRUCTURE: Define the broken profile as the boundary between processed and unprocessed cells. State the number of plug positions for an n-row grid. Describe the geometric shape of this boundary in terms of horizontal and vertical segments.

2. STATE BOUND: Using Catalan numbers, state the upper bound on the number of distinct canonical plug states for a profile with (n+1) positions. Give the exact numeric value for n = 12.

3. TRANSITION RULES FOR EMPTY CELLS: When processing an empty cell (i, j), define all valid transitions based on the values of the LEFT plug and TOP plug:
   (a) Both LEFT = 0 and TOP = 0
   (b) Exactly one of LEFT or TOP is nonzero
   (c) Both nonzero with different component labels
   (d) Both nonzero with the same component label
   For each case, state what outgoing plug values are produced and any validity constraints.

4. PREMATURE CYCLE REJECTION: Explain what happens in case (d) at a non-terminal cell and why the transition must be handled differently than case (c).

5. COMPONENT MERGING: When two components with different labels are merged in case (c), specify how the plug array is updated and how re-canonicalization is performed.

6. START CELL (0,0): State the valid outgoing plug configuration(s) for the path's starting endpoint. Specify which outgoing plugs may be nonzero and which must be zero, and justify using the degree constraint.

7. END CELL (n-1, m-1): State the required incoming plug configuration for valid Hamiltonian path termination. Specify conditions on LEFT, TOP, and all other plug positions.

8. SPACE MANAGEMENT: Describe how DP states are stored across cell transitions. State how many collections of states must be maintained simultaneously and the resulting space complexity in terms of Catalan numbers.

9. DP RECURRENCE: Write the DP update rule relating dp_next and dp_curr, specifying where canonicalization is applied.

10. TIME AND SPACE COMPLEXITY: State the overall time and space complexity in terms of n, m, and Catalan numbers. Justify feasibility for n = 12, m = 1000 by citing the numeric Catalan bound.

For verification: on a 3x3 grid with all cells empty, the number of Hamiltonian paths from (0,0) to (2,2) is exactly 2.

The response states that the broken profile has exactly n+1 plug positions for an n-row grid, and describes the profile as consisting of n horizontal segments and 1 vertical segment (or equivalently as a staircase or L-shaped boundary).

The response states that the number of distinct canonical plug states is bounded by the Catalan number C_{n+1}, and provides the exact numeric value C_13 = 742900 as the upper bound when n = 12.

The response defines canonicalization as assigning component labels 1, 2, 3, ... in left-to-right first-occurrence order across the plug array, ensuring a unique representation for each distinct connectivity pattern.

The response correctly states that when both LEFT and TOP plugs carry the same component label at a non-terminal cell, the transition must be rejected because it would create a premature cycle, which violates the Hamiltonian path property.

The response specifies that when merging two components with different labels, all occurrences of one label are replaced with the other throughout the plug array, followed by re-canonicalization of the entire state.

The response states that at the start cell (0,0), exactly one outgoing plug is set to a nonzero component label and the other must remain 0, because a path endpoint has degree 1.

The response states that at the end cell (n-1, m-1), exactly one of the two incoming plugs (LEFT or TOP) must be nonzero while the other is 0, AND all remaining plug positions in the array must be exactly 0.

The response provides the DP recurrence as dp_next[canonicalize(new_state)] += dp_curr[old_state] for each valid transition, or an equivalent formulation that explicitly applies canonicalization to the new state before accumulation.

The response states that exactly two DP state collections (current and next) are maintained at any time, and that this achieves space complexity of O(C_{n+1}) rather than storing states for all cells.

The response states the time complexity as O(n * m * C_{n+1}) or an asymptotically equivalent expression, and confirms feasibility for n = 12 and m = 1000 by citing the numeric bound C_13 = 742900.

1. PROFILE STRUCTURE:
The broken profile is the boundary between processed cells (above and to the left of the current cell in column-major order) and unprocessed cells (below and to the right). It forms a staircase or L-shaped boundary consisting of exactly n horizontal segments and 1 vertical segment. The profile has exactly n+1 plug positions for a grid with n rows.

2. STATE BOUND:
The number of distinct canonical plug states on (n+1) positions is bounded by the Catalan number C_{n+1}. The Catalan number C_k = (2k)! / ((k+1)! * k!). For n = 12, the profile has 13 positions, so the bound is C_13 = 742,900.

3. TRANSITION RULES FOR EMPTY CELLS:
When processing cell (i, j), let L = LEFT plug value and T = TOP plug value.

(a) L = 0 and T = 0:
For an interior cell (not start or end): the cell must have degree 2, so if no segment enters from LEFT or TOP, the only valid choice is to start a new open component at this cell. Create a new component label k and set both outgoing plugs to that label: RIGHT = k and DOWN = k. This means the current cell contributes two incident edges, one to the right and one downward, and those two dangling ends belong to the same component to be connected later.
For the start or end cell: degree is 1, so exactly one outgoing plug gets label k, the other stays 0 (see points 6 and 7).

(b) Exactly one of L, T is nonzero (say value = p, the other = 0):
The existing segment continues through this cell. Two sub-options for degree-2 cells:
  - Continue RIGHT: set RIGHT plug = p, DOWN plug = 0
  - Continue DOWN: set DOWN plug = p, RIGHT plug = 0
For an endpoint cell (degree 1): the segment terminates. Set both outgoing plugs to 0.

(c) Both L and T nonzero with different labels (L = p, T = q, p != q):
The cell merges two distinct path segments. Both outgoing plugs become 0 (the cell consumes both incoming segments). The two components must be unified: replace all occurrences of max(p, q) with min(p, q) throughout the plug array, then re-canonicalize (see point 5).

(d) Both L and T nonzero with the same label (L = T = p):
This would close a cycle, so it is invalid for a Hamiltonian path and must be rejected. See point 4.

4. PREMATURE CYCLE REJECTION:
At any non-terminal cell, if L and T have the same component label, connecting them would join two ends of the same partial component and immediately create a cycle before all cells are finished. That is a premature cycle, so the transition must be rejected.

The terminal cell is also rejected in case (d), but for the full endpoint condition: using both incoming plugs would give degree 2 at (n-1, m-1), whereas a Hamiltonian path ending there requires degree 1. So valid termination is only the one-incoming-plug case from point 7.

5. COMPONENT MERGING:
When merging labels p and q (assume p < q without loss of generality): scan the entire plug array of (n+1) entries and replace every occurrence of q with p. Then re-canonicalize: scan the array left-to-right and assign new labels 1, 2, 3, ... to each distinct nonzero value in order of first appearance. This ensures the resulting state is in canonical form for hash map lookup.

6. START CELL (0,0):
Cell (0,0) is a path endpoint and must have degree 1. Since it is at position (0,0), there is no LEFT plug and no TOP plug (both are 0 by boundary conditions). The cell must produce exactly one outgoing plug set to component label 1:
  - Option A: RIGHT plug = 1, DOWN plug = 0 (path starts going right)
  - Option B: RIGHT plug = 0, DOWN plug = 1 (path starts going down)
Setting both to 1 would give degree 2, violating the endpoint constraint.

7. END CELL (n-1, m-1):
Cell (n-1, m-1) is the other path endpoint with degree 1. For a valid Hamiltonian path termination:
  - Exactly one of LEFT or TOP must be nonzero (one incoming segment)
  - The other must be zero
  - All remaining (n-1) plug positions in the array must be exactly 0
This ensures that no dangling path segments exist elsewhere in the grid, meaning all free cells have been incorporated into a single path from (0,0) to (n-1, m-1).

8. SPACE MANAGEMENT:
Two hash maps (or dictionaries) are maintained: dp_curr (states for the current cell being processed) and dp_next (states being built for the next cell). For each cell, iterate over all entries in dp_curr, compute valid transitions, and accumulate into dp_next. After processing the cell, set dp_curr = dp_next and reset dp_next to empty. This alternation means only O(C_{n+1}) state entries are stored at any time, giving space complexity O(C_{n+1}).

9. DP RECURRENCE:
For each state old_state in dp_curr with value dp_curr[old_state]:
  For each valid transition at the current cell producing new_state:
    dp_next[canonicalize(new_state)] += dp_curr[old_state]  (mod 10^9 + 7)

Here canonicalize(new_state) relabels the component IDs in the plug array as 1, 2, 3, ... by left-to-right first occurrence, ensuring a unique key for each connectivity pattern.

10. TIME AND SPACE COMPLEXITY:
- Number of distinct DP states per cell: O(C_{n+1}), bounded by Catalan number C_{n+1}
- For n = 12: C_13 = 742,900 states
- Transitions per state per cell: O(1) amortized (constant number of cases, plus O(n) for re-canonicalization, but n <= 12 is treated as constant)
- Total cells processed: n * m
- Time complexity: O(n * m * C_{n+1})
- Space complexity: O(C_{n+1})
- Feasibility for n = 12, m = 1000: the state cap is C_13 = 742,900, so memory remains bounded by storing at most two such state collections at once. This Catalan bound is the key reason the method is viable for width 12; in practice the number of reachable states is typically well below the worst-case cap.

VERIFICATION:
For the 3x3 all-empty grid, the two Hamiltonian paths from (0,0) to (2,2) are:
  Path 1: (0,0) -> (0,1) -> (0,2) -> (1,2) -> (1,1) -> (1,0) -> (2,0) -> (2,1) -> (2,2)
  Path 2: (0,0) -> (1,0) -> (2,0) -> (2,1) -> (1,1) -> (0,1) -> (0,2) -> (1,2) -> (2,2)
Count = 2.


Prompt:
You are fighting a vertical column of `n` enemies arranged in a single stack.

- Enemy `1` is at the bottom, enemy `n` is at the top.
- Enemy `i` starts with health `h[i]`.

### Attack rule
In one hit, you may choose **any currently alive enemy** (in any stack that exists at that moment) and reduce its health by `1`.

### Collapse rule after a death
Whenever an enemy’s health becomes `0` or less, that enemy dies immediately. Then:

1. All enemies that were **above** the dead one fall downward and become a **new stack** (separate from whatever remains below).
2. In this new stack, the **bottom** enemy takes **fall damage** equal to how many enemies were below it before it fell.
   That number includes the enemy that just died.
3. If the bottom enemy of the new stack dies from fall damage, the same falling process repeats again for the enemies above it, potentially causing a chain reaction.

This continues until no newly formed stack has its bottom enemy killed by the fall damage it just received.

### Goal
Determine the minimum number of hits required to kill **all** enemies.

---

## Input
The input contains multiple test cases.

- The first line contains an integer `t` (`1 ≤ t ≤ 10^4`).
- For each test case:
  - The first line contains an integer `n` (`2 ≤ n ≤ 2·10^5`) — the number of enemies in the initial stack.
  - The second line contains `n` integers `h[1], h[2], …, h[n]` (`1 ≤ h[i] ≤ 10^9`) — their initial health values.

It is guaranteed that the sum of `n` over all test cases does not exceed `2·10^5`.

---

## Output
For each test case, output one integer: the minimum number of hits needed to eliminate every enemy.

Does the answer define dp[i] as the minimum number of attacks needed to eliminate mobs 1 through i?

Does the answer state that the final output for a test case equals dp[n]?

Does the answer state that a mob can receive fall damage at most once?

Does the answer state the base case dp[0] = 0?

Does the answer state the base case dp[1] = h[1]?

Does the answer state that one option makes mob i take exactly 1 fall damage?

Does the answer give the transition term dp[i-1] + max(0, h[i] - 1)?

Does the answer state that the other option makes mob i take exactly i-1 fall damage?

Does the answer give the transition term dp[i-2] + h[i-1] + max(0, h[i] - (i-1))?

Does the answer state a per-test-case time complexity of O(n)?

### Key observations
1. **A creature can receive fall damage at most once.**
   After it falls, it becomes the bottom of its new stack, and nothing can later fall “under it” again.

2. **If a creature ever receives fall damage greater than 1,** then the creature directly below it must have been killed by sword hits (not by fall damage).
   Reason: a creature that has already fallen is at the bottom; killing it would only ever make the next creature fall by **1**.

3. Therefore, whenever we want a creature `i` to get a large fall damage amount, we should arrange it so that creature `i-1` is killed **while the original stack below is still intact**, because that maximizes the fall damage to `i` without increasing the number of sword hits needed elsewhere.

---

### DP definition
Let `dp[i]` be the minimum number of sword hits needed to eliminate creatures `1..i` (from the bottom up, considering only these `i` creatures).

Base:
- `dp[0] = 0`
- `dp[1] = h[1]`

For `i ≥ 2`, we consider two ways to handle creature `i`:

---

### Option A: creature `i` only falls by 1
If creature `i-1` dies when it is the bottom of the current stack containing `i-1` and `i`, then `i` falls and takes **1** damage.

Cost:
- clear `1..i-1`: `dp[i-1]`
- finish creature `i` after 1 fall damage: `max(0, h[i] - 1)`

So:
\[
\text{costA} = dp[i-1] + \max(0, h[i] - 1)
\]

---

### Option B: creature `i` gets maximum fall damage `(i-1)`
Kill creature `i-1` directly **before** removing any of `1..i-2`, so creature `i` falls from the original height and takes `(i-1)` damage.

Cost:
- clear `1..i-2` eventually: `dp[i-2]`
- kill creature `i-1` by hits: `h[i-1]`
- finish creature `i` after `(i-1)` fall damage: `max(0, h[i] - (i-1))`

So:
\[
\text{costB} = dp[i-2] + h[i-1] + \max(0, h[i] - (i-1))
\]

---

### Transition and answer
\[
dp[i] = \min(\text{costA}, \text{costB})
\]
The required minimum hits for the full stack is `dp[n]`.

---

### Complexity
Each `dp[i]` is computed in `O(1)`, so the solution is `O(n)` per test case (and `O(\sum n)` overall).


Prompt:
You are given a tree with `n` nodes. Node `i` has an allowed initial value interval `[l_i, r_i]`.
You choose initial values `a_i` such that:
`l_i ≤ a_i ≤ r_i`

A tree is **balanced** if all node values are equal; the balanced value is that common value.

**Operation:** choose two nodes `u` and `v`. Consider the tree rooted at `u`. Increase by `1` the values of all nodes in the subtree of `v` (in that rooting). You may choose `u = v`. Values can only increase.

Find the **minimum possible** final balanced value after any number of operations.

### Input
- `n` (`1 ≤ n ≤ 2·10^5`)
- `n` lines: `l_i, r_i` (`0 ≤ l_i ≤ r_i ≤ 10^9`)
- `n-1` edges forming a tree

### Output
One integer: the minimum achievable balanced value.

Does the answer state that for an edge between parent p and child v (in a chosen rooting), choosing (u=p,v=v) increases (a_v−a_p) by 1 and choosing (u=v,v=p) decreases (a_v−a_p) by 1 (i.e., one operation changes the edge difference by exactly ±1)?

Does the answer state that for fixed initial values a, making a_v = a_p on an edge (p,v) requires exactly |a_v−a_p| operations affecting that edge (because each such operation changes the difference by 1)?

Does the answer give the final balanced value for a fixed rooting at 1 as F = a_1 + sum_{v≠1}?

Does the answer define the bottom-up choice of initial values as a_v = min(r_v, max(l_v, max_{child u of v}))?

Does the answer justify that if r_v allows a_v ≥ max_child, then choosing a_v below max_child cannot improve the objective (it only increases required increments on edges to children)?

Does the answer explicitly handle the clipped case: if r_v < max_child, then a_v is set to r_v and the deficit (max_child − r_v) is necessarily paid later via the term max(0, a_child − a_v)?

Does the answer describe computing the chosen a_v values using a postorder DFS from root 1 (children processed before parent)?

Does the answer describe computing the final answer by accumulating sum_{v≠1}?

Does the answer state an implementation complexity of O(n) time and O(n) memory and mention using 64-bit integers for the answer?

### Expert Solution

Root the tree at node `1`. Let `p(v)` be the parent of `v`.

---

### A) Final balanced value for fixed initial values `a`

Consider an edge `(p, v)`.

An operation “increment subtree of `v` when rooted at `u`” can be used in two relevant ways for this edge:

- Choose `u = p`, `v = v`: increments the `v`-side component, so `(a_v - a_p)` **increases by 1**.
- Choose `u = v`, `v = p`: increments the `p`-side component (the side containing node `1`), so `(a_v - a_p)` **decreases by 1**.

So each operation changes the edge difference by exactly `±1`, hence equalizing the edge requires exactly `|a_v - a_p|` operations affecting it.

If `a_v > a_p`, to make them equal we must decrease `(a_v - a_p)` to `0`, which forces incrementing the **root side** exactly `a_v - a_p` times, increasing the final common value by that amount.
If `a_v ≤ a_p`, we can equalize by incrementing only the `v`-side, which does **not** increase the root-side final value.

Therefore the final balanced value (with root fixed at `1`) is:
`F(a) = a_1 + ∑_{v ≠ 1} max(0, a_v - a_{p(v)})`

---

### B) Optimal choice of initial values `a_v ∈ [l_v, r_v]`

Compute `a` bottom-up. For node `v`, let
`M_v = max_{u child of v} a_u`
(with `M_v = -∞` for a leaf).

If `r_v` allows `a_v ≥ M_v`, choosing `a_v < M_v` only increases the required increments on edges to children (since some child would exceed `v` more), so it cannot reduce `F(a)`. Thus we pick:
`a_v = min(r_v, max(l_v, M_v))`

If `r_v < M_v`, we are forced to set `a_v = r_v`; the unavoidable deficit is then paid later via terms `max(0, a_u - a_v)` for the too-large children.

---

### Algorithm

1. Root at `1` and run a postorder DFS to compute
   `a_v = min(r_v, max(l_v, max_{child} a_{child}))`

2. Compute
   `ans = a_1 + ∑_{v ≠ 1} max(0, a_v - a_{p(v)})`

Use 64-bit integers.

---

### Complexity

`O(n) time`, `O(n) memory`
which fits `n ≤ 2·10^5`.


Prompt:
## Problem: Eliminate “+1 Later” Conflicts

You are given an integer sequence `a` of length `n`. Call a sequence `b` **valid** if there are **no** positions `i < j` such that:

\[
b_j = b_i + 1
\]

In other words, you must avoid any situation where a value appears and **somewhere later** a value exactly **one larger** appears.

You may delete elements from `a` (keeping the remaining elements in their original order). Your task is to compute the **minimum number of deletions** required so that the remaining sequence becomes valid.

---

## Input
The input contains multiple test cases.

- The first line contains an integer `t` (`1 ≤ t ≤ 6·10^4`) — the number of test cases.
- For each test case:
  - The first line contains an integer `n` (`1 ≤ n ≤ 3·10^5`) — the length of the sequence.
  - The second line contains `n` integers `a1, a2, …, an` (`1 ≤ ai ≤ n`).

It is guaranteed that the sum of `n` over all test cases does not exceed `3·10^5`.

---

## Output
For each test case, output one integer: the minimum number of elements that must be removed from `a` so that the remaining sequence is valid.

Does the answer state the forbidden pattern as existence of indices i<j with b[j] = b[i] + 1?

Does the answer sort the pair list by value in non-decreasing order?

Does the answer define dp[i] as the maximum size of a good selection whose last chosen pair is the i-th pair in the sorted list?

Does the answer include a maintained maximum over dp[j] for processed pairs satisfying value[j] ≤ value[i] − 2?

Does the answer include a query for the maximum dp among processed pairs satisfying (value = current value − 1, position > current position)?

Does the answer state that the final result equals n minus the maximum attainable kept length (answer = n − keep)?

Does the answer construct a list of pairs (a[i], i) using the original index as the second component?

Does the answer specify that equal values are tie-broken by sorting indices in non-increasing (descending) order?

Does the answer specify the disallowed transition condition (value[i] = value[j] + 1, pos[i] > pos[j])?

Does the answer include a maintained maximum dp among processed pairs with value equal to the current value?

### Key reformulation
We want to keep as many elements as possible. Let `keep` be the maximum size of a subsequence that is **good**; then the answer is:

\[
\text{removed} = n - keep
\]

A subsequence is **not good** exactly when it contains two chosen indices `p<q` with `a[q]=a[p]+1`.
So, if we choose some elements of value `x` and some of value `x+1`, then **every chosen `(x+1)` must lie strictly to the left of every chosen `x`**.

---

### Step 1: Sort values with positions
Create pairs `(value, position)` for all indices.

Sort these pairs by:
1. `value` increasing
2. `position` decreasing for equal values

Call the sorted list `v[1..n]`, where `v[i] = (val[i], pos[i])`.

This tie-break (positions descending inside equal values) will let us treat “taking multiple equal values” cleanly while enforcing constraints only between consecutive values.

---

### Step 2: DP definition
Define:

- `dp[i]` = maximum number of kept elements in a good subsequence **whose last selected pair is exactly** `v[i]`.

Then:

\[
dp[i] = 1 + \max dp[j]
\]
over all `j < i` that can precede `i` without creating a forbidden `(+1 later)` pair.

When is transition `j -> i` invalid?
- It is invalid only in the specific case:
  - `val[i] = val[j] + 1` **and**
  - `pos[i] > pos[j]`

Because that would mean an element of value `val[j]` occurs earlier in the kept subsequence and a `+1` occurs later, which is forbidden.

So the transition is allowed iff:

\[
val[i] \ne val[j] + 1 \quad \text{or} \quad pos[i] < pos[j]
\]

Finally:

\[
keep = \max_{1 \le i \le n} dp[i],
\quad \text{answer} = n - keep
\]

---

### Step 3: Speeding up the transition
A direct computation is `O(n^2)`, too slow. We optimize by splitting candidates `j` into three disjoint groups:

#### A) Values at most `val[i]-2`
If `val[j] <= val[i]-2`, then `val[i]` cannot equal `val[j]+1`, so the transition is always valid.

Maintain:
- `bestSmall` = maximum `dp[j]` among all processed `j` with `val[j] <= currentValue-2`.

#### B) Same value: `val[j] = val[i]`
Always valid (since the forbidden condition requires `+1` difference).

Maintain:
- `bestSame[value]` = maximum `dp[j]` seen so far for this exact value.

#### C) Previous value: `val[j] = val[i]-1`
This is the only case where position matters. We need `pos[i] < pos[j]`.

Inside each value group, positions are processed in **decreasing** order (because of the sort).
So for value `val[i]-1`, we can maintain a structure that answers:
- maximum `dp[j]` among those earlier in the group with `pos[j] > pos[i]`

This can be done with a running **suffix maximum over the group** (equivalently: as you sweep positions downward, maintain the best dp seen so far).

Call this query result:
- `bestPrevGreaterPos`

Then:

\[
dp[i] = 1 + \max\big(bestSmall,\; bestSame[val[i]],\; bestPrevGreaterPos\big)
\]

After computing `dp[i]`, update the maintained maxima for future steps.

---

### Complexity
- Sorting pairs: `O(n log n)` per test file (overall across test cases).
- DP with maintained maxima: `O(n)` after sorting.
- Total across all tests: `O((\sum n)\log(\sum n))`, with `\sum n \le 3\cdot 10^5`.

This meets the constraints.

---