Prompt:
A database system executes concurrent transactions under Snapshot Isolation (SI). The system must detect write-skew anomalies and compute conflict metrics.

CRITICAL CLARIFICATION: All reads occur instantaneously at the transaction's start time (SI snapshot semantics).

Transaction Schedule (10 transactions):
T1: R(X)10 R(Y)20 W(X)30
T2: R(Y)20 R(Z)15 W(Y)40
T3: R(X)10 R(Z)15 W(Z)25
T4: R(Z)25 R(Y)40 W(X)50
T5: R(X)30 R(Y)40 W(Y)60
T6: R(Y)60 R(Z)25 W(Z)35
T7: R(Z)35 R(X)50 W(X)70
T8: R(X)70 R(Y)60 W(Y)80
T9: R(Y)80 R(Z)35 W(Z)45
T10: R(Z)45 R(X)70 W(X)90

Notation: R(X)v means "Read variable X with value v", W(X)v means "Write variable X with value v"

Execution Timeline (timestamp order):
T1: [0,15], T2: [5,20], T3: [10,25], T4: [12,30], T5: [18,35]
T6: [22,40], T7: [28,45], T8: [33,50], T9: [38,55], T10: [42,60]

Format: [start_time, commit_time] in milliseconds

Conflict Detection Rules:

1. Write-Read (WR) Conflict: Ti writes X at value v, Tj reads X with value u where u ≠ v, and Ti commits before Tj starts
2. Read-Write (RW) Conflict: Ti reads X with value v, Tj writes X with value u where u ≠ v, and start_time(Ti) ∈ [start_time(Tj), commit_time(Tj)]
3. Write-Write (WW) Conflict: Ti writes X at value v, Tj writes X with value u where u ≠ v, and their execution windows overlap

Analysis Requirements:

Phase 1: Conflict Graph Construction
For each conflict type (WR, RW, WW), identify all pairs (Ti, Tj) that satisfy the conflict condition.

Phase 2: Conflict Metric Calculation
For each detected conflict (Ti, Tj):

Step 1: Compute Temporal Distance (TD):
TD = |commit_time(Ti) - start_time(Tj)|

Step 2: Compute Value Divergence (VD):
- Let Vi be the set of all values written by Ti
- Let Vj be the set of all values written by Tj
- VD = Σ|vi - vj| for all pairs (vi, vj) where vi ∈ Vi and vj ∈ Vj

Step 3: Compute Conflict Hash (CH):
CH = (i × 1000 + j × 100 + TD) mod 997

Step 4: Compute Prime Factor Sum (PFS):
- Factorize CH into prime factors
- If CH is prime, PFS = CH
- Otherwise, PFS = sum of all prime factors (with repetition)

Step 5: Update Cumulative Conflict Score (CCS):
CCS = CCS + (PFS × VD / 100)
Initially, CCS = 0.

Phase 3: Serializability Analysis
A cycle exists in the conflict graph if there is a path Ti → Tj → ... → Tk → Ti.
Count all simple cycles (no repeated transactions except start/end).

Required Outputs:

1. Total_WR_Conflicts: Count of Write-Read conflicts
2. Total_RW_Conflicts: Count of Read-Write conflicts
3. Total_WW_Conflicts: Count of Write-Write conflicts
4. Final_CCS: Cumulative Conflict Score rounded to 2 decimal places
5. Total_Cycles: Number of simple cycles in the conflict graph
6. Largest_Cycle_Length: Length (number of transactions) in the longest cycle
7. Is_Serializable: "YES" if no cycles exist, "NO" otherwise

Provide your answer in exactly this format:
Total_WR_Conflicts: [integer]
Total_RW_Conflicts: [integer]
Total_WW_Conflicts: [integer]
Final_CCS: [decimal with 2 places]
Total_Cycles: [integer]
Largest_Cycle_Length: [integer]
Is_Serializable: [YES/NO]

Does the answer treat all reads as instantaneous events occurring strictly at start_time(Ti) and not over the transaction’s execution window?

Does the answer conclude Total_WR_Conflicts = 0 by explicitly applying the condition that a WR conflict requires u ≠ v under snapshot isolation?

Does the answer identify RW conflicts only when start_time(Ti) ∈ [start_time(Tj), commit_time(Tj)] and not based on transaction lifetime overlap?

Does the answer correctly compute Value Divergence (VD) for each conflict as the absolute difference between the values written by Ti and Tj?

Does the answer correctly compute and report Final_CCS as 132.48 after summing all conflict contributions?

Does the answer compute Temporal Distance using the exact formula TD = |commit_time(Ti) − start_time(Tj)| for every conflict?

Does the answer correctly report Total_Cycles = 0 and Largest_Cycle_Length = 0 based on the constructed graph?

Does the answer identify exactly 7 Read–Write (RW) conflicts, neither omitting nor adding any pairs?

Does the answer correctly report Is_Serializable as YES based on zero detected cycles?

## 1. Conflict Identification (Phase 1)

### 1.1 Write–Read (WR) Conflicts
Under Snapshot Isolation, each transaction reads the most recent committed value strictly before its start time.

For a Write–Read (WR) conflict to exist, there must be a pair (Ti, Tj) such that:

• Ti writes X = v
• Ti commits before start_time(Tj)
• Tj reads X = u
• and u ≠ v

For every read operation in the schedule, the value read equals the value written by the last transaction that committed before the reader’s start time.

Therefore, there is no pair (Ti, Tj) satisfying the value inequality condition (u ≠ v), and no WR conflicts occur.

### 1.2 Read–Write (RW) Conflicts

A Read–Write (RW) conflict exists if transaction (T_i) reads variable (X) at its start time, transaction (T_j) writes the same variable with a different value, and

start_time(T_i) ∈ [start_time(T_j), commit_time(T_j)].

The conflict edge direction is Ti → Tj (reader → writer).

Applying this rule strictly:

1. (T_4) reads (Z) at time 12 while (T_3) executes from 10 to 25 → (T_4 --> T_3)
2. (T_5) reads (Y) at time 18 while (T_2) executes from 5 to 20 → (T_5 --> T_2)
3. (T_6) reads (Z) at time 22 while (T_3) executes from 10 to 25 → (T_6 --> T_3)
4. (T_7) reads (X) at time 28 while (T_4) executes from 12 to 30 → (T_7 --> T_4)
5. (T_8) reads (Y) at time 33 while (T_5) executes from 18 to 35 → (T_8 --> T_5)
6. (T_9) reads (Z) at time 38 while (T_6) executes from 22 to 40 → (T_9 --> T_6)
7. (T_10) reads (X) at time 42 while (T_7) executes from 28 to 45 → (T_10 --> T_7)

Result:
Total_RW_Conflicts = 7

### 1.3 Write–Write (WW) Conflicts

A Write–Write (WW) conflict exists if two transactions write the same variable with different values and their execution windows overlap. The edge direction is from the earlier writer to the later writer.

Identified conflicts:

Variable X

1. (T_1 --> T_4) (overlap [12,15])
2. (T_4 --> T_7) (overlap [28,30])
3. (T_7 --> T_10) (overlap [42,45])

Variable Y
4. (T_2 --> T_5) (overlap [18,20])
5. (T_5 --> T_8) (overlap [33,35])

Variable Z
6. (T_3 --> T_6) (overlap [22,25])
7. (T_6 --> T_9) (overlap [38,40])

Result:
Total_WW_Conflicts = 7

## 2. Conflict Metric Calculation (Phase 2)

Total conflicts:

0(WR) + 7(RW) + 7(WW) = 14

Each transaction writes exactly one value, so the Value Divergence (VD) is the absolute difference between the two written values.

### 2.1 Example RW Conflict: (T_4 --> T_3)

 Temporal Distance
TD = |commit_time(Ti) − start_time(Tj)|
   = |commit_time(T4) − start_time(T3)|
   = |30 − 10| = 20
 Value Divergence (VD): (|50 - 25| = 25)
 Conflict Hash (CH):
  ((4 x 1000 + 3 x 100 + 20) mod 997 = 332)
 Prime Factor Sum (PFS):
  (332 = 2^2 x 83 --> 2 + 2 + 83 = 87)
 CCS Increment:
  ((87 x 25)/100 = 21.75)

### 2.2 Example WW Conflict: (T_1 --> T_4)

 TD: (|15 - 12| = 3)
 VD: (|30 - 50| = 20)
 CH:
  ((1 x 1000 + 4 x 100 + 3) mod 997 = 406)
 PFS:
  (406 = 2 x 7 x 29 --> 38)
 CCS Increment:
  ((38 x 20)/100 = 7.60)

### 2.3 Final CCS

Summing the CCS increments for all 14 conflicts:

Final_CCS = 132.48

## 3. Serializability Analysis (Phase 3)

### 3.1 Conflict Graph Edges

WW edges (writer → writer):

 (T_1 --> T_4 --> T_7 --> T_10)
 (T_2 --> T_5 --> T_8)
 (T_3 --> T_6 --> T_9)

RW edges (reader → writer):

 (T_4 --> T_3)
 (T_5 --> T_2)
 (T_6 --> T_3)
 (T_7 --> T_4)
 (T_8 --> T_5)
 (T_9 --> T_6)
 (T_10 --> T_7)

### 3.2 Cycle Detection

No directed path returns to its starting transaction. Therefore, the conflict graph contains no simple cycles.

Results:

 Total_Cycles = 0
 Largest_Cycle_Length = 0
 Is_Serializable = YES

## 4. Final Answer (Copy-Paste Ready)

Total_WR_Conflicts: 0
Total_RW_Conflicts: 7
Total_WW_Conflicts: 7
Final_CCS: 132.48
Total_Cycles: 0
Largest_Cycle_Length: 0
Is_Serializable: YES


Prompt:
A processor allows instructions to overlap in execution by using a centralized scoreboarding mechanism rather than strict in-order completion. Execution resources are limited and non-pipelined: there is exactly one Integer execution unit (1-cycle latency), one floating-point unit capable of addition and subtraction (4-cycle latency), one floating-point multiplier (6-cycle latency), and one floating-point divider (9-cycle latency). No functional unit may accept a new instruction until the current one has fully completed execution.

Each instruction progresses through four pipeline stages: dispatch, operand acquisition, execution, and result commitment. Dispatch is permitted only if the required functional unit is unused. Operand acquisition may not proceed until all required source operands are produced, thereby enforcing RAW correctness. Result commitment may be delayed if committing the value would violate a WAR dependency. Load instructions obtain their data using the Integer execution unit and follow identical hazard rules.

At the beginning of cycle 1, the register file contains the following values:
R3 = 12, R5 = 24
F1 = 1.5, F3 = 3.0, F5 = 5.0, F7 = 7.0, F9 = 9.0, F11 = 11.0

Starting at cycle 1, the processor encounters the following instruction sequence in program order:

I1: Load F3 from the memory location addressed by R3
I2: Multiply F5 using operands F3 and F7
I3: Add F7 using operands F5 and F9
I4: Divide F9 using operands F7 and F3
I5: Subtract to produce F11 using F9 and F5
I6: Load F1 from the memory location addressed by R5
I7: Multiply F3 using operands F1 and F11
I8: Add F5 using operands F3 and F7
I9: Divide F7 using operands F5 and F1
I10: Subtract to produce F9 using F7 and F3

For each instruction, determine exactly which cycle it is dispatched, successfully reads its operands, completes execution and commits its result to the register file.

Using this execution timeline, compute the following quantities:

1. The cycle at which the final instruction finishes committing its result
2. The total of all dispatch cycle numbers
3. The total of all result-commit cycle numbers
4. The cumulative number of stall cycles caused by WAW conflicts
5. The commit cycle of instruction I6
6. The earliest cycle at which the integer execution unit becomes permanently idle
7. The maximum number of instructions simultaneously present in the system (issued but not yet committed)
8. The instruction that experiences the longest continuous stall period
9. The total number of cycles during which no instruction is able to dispatch
10. The average execution-unit utilization across all functional units over the entire timeline

Does the answer compute the cycle at which the final instruction finishes committing to be exactly cycle 66?

Does the answer compute the total of all dispatch cycle numbers to be exactly 193?

Does the answer compute the total of all result-commit cycle numbers to be exactly 339?

Does the answer compute the cumulative number of stall cycles caused by WAW conflicts to be exactly 0?

Does the answer state the commit cycle of instruction I6 to be cycle 23?

Does the answer state the earliest cycle at which the integer execution unit becomes permanently idle to be cycle 24?

Does the answer compute the maximum number of instructions simultaneously present in the system to be exactly 5?

Does the answer state the instruction that experiences the longest continuous stall period to be exactly I4?

Does the answer compute the total number of cycles during which no instruction is able to dispatch to be exactly 56?

Does the answer compute the average execution-unit utilization across all functional units over the entire timeline to be approximately 18.2%?

Cycle 1:
- I1: Load F3 from the memory location addressed by R3
        - Integer unit is free
        - I1 dispatches
- End state:
        - I1 dispatched
        - Integer unit reserved by I1, rest are free
        - F3 pending. Other registers unchanged.

Cycle 2:
- I1: Load F3 from the memory location addressed by R3
        - R3 ready
        - I1 successfully acquire operands
- I2: Multiply F5 using operands F3 and F7
        - Multiplier unit free
        - I2 dispatches
- End state:
        - I1 successfully acquire operands, I2 dispatched
        - Integer unit reserved by I1, Multiplier unit reserved by I2, rest are free
        - F3(I1) and F5(I2) pending. Other registers unchanged.

Cycle 3:
- I1: Load F3 from the memory location addressed by R3
        - Begins execution
        - Will execute for 1 cycle
        - Completes execution by end of cycle 3
- I2: Multiply F5 using operands F3 and F7
        - F7 is free but F3 is pending
        - Operand read stalls
- I3: Add F7 using operands F5 and F9
        - Adder unit free
        - I3 dispatches
- End state:
        - I1 executes for complete cycle, I2 stalled in Operand Read (RAW on F3), I3 dispatched
        - Integer unit reserved by I1, Multiplier unit reserved by I2, Adder unit reserved by I3, Divider free
        - F3(I1), F5(I2) and F7(I3) pending. Other registers unchanged.

Cycle 4:
- I1: Load F3 from the memory location addressed by R3
        - Commits F3
        - Leaves the system
- I2: Multiply F5 using operands F3 and F7
        - F7 is free but F3 is pending
        - Operand read stalls
- I3: Add F7 using operands F5 and F9
        - F9 available but F5 is pending
        - Operand read stalls
- I4: Divide F9 using operands F7 and F3
        - Divider unit free
        - I4 dispatches
- End state:
        - I1 commits F3 and leaves, I2 stalled in Operand Read (RAW on F3), I3 stalled in Operand Read (RAW on F5), I4 dispatched
        - Integer unit free, Multiplier unit reserved by I2, Adder unit reserved by I3, Divider unit reserved by I4
        - F5(I2), F7(I3) and F9(I4) pending. F3 ready. Other registers unchanged.

Cycle 5:
- I2: Multiply F5 using operands F3 and F7
        - F3 and F7 both available (F7 is pending by later instruction)
        - Successfully acquire operands
- I3: Add F7 using operands F5 and F9
        - F9 available but F5 is pending
        - Operand read stalls
- I4: Divide F9 using operands F7 and F3
        - F3 is available but F7 is pending by I3
        - Operand read stalls
- I5: Subtract to produce F11 using F9 and F5
        - Adder not available
- End state:
        - I2 successfully acquire operands, I3 stalled in Operand Read (RAW on F5), I4 stalled in Operand Read (RAW on F7), I5 cannot dispatch
        - Integer unit free, Multiplier unit reserved by I2, Adder unit reserved by I3, Divider unit reserved by I4
        - F5(I2), F7(I3) and F9(I4) pending. F3 ready. Other registers unchanged.

Cycle 6:
- I2: Multiply F5 using operands F3 and F7
        - Begins execution
        - Will execute for 6 cycles
        - Will complete execution by cycle 11
- I3: Add F7 using operands F5 and F9
        - F9 available but F5 is pending
        - Operand read stalls
- I4: Divide F9 using operands F7 and F3
        - F3 is available but F7 is pending by I3
        - Operand read stalls
- I5: Subtract to produce F11 using F9 and F5
        - Adder not available
- End state:
        - I2 starts execution, I3 stalled in Operand Read (RAW on F5), I4 stalled in Operand Read (RAW on F7), I5 cannot dispatch
        - Integer unit free, Multiplier unit reserved by I2, Adder unit reserved by I3, Divider unit reserved by I4
        - F5(I2), F7(I3) and F9(I4) pending. F3 ready. Other registers unchanged.

Cycle 7-11:
- I2: Multiply F5 using operands F3 and F7
        - Executing (Finishes by the end of cycle 11)
- I3: Add F7 using operands F5 and F9
        - F9 available but F5 is pending
        - Operand read stalls
- I4: Divide F9 using operands F7 and F3
        - F3 is available but F7 is pending by I3
        - Operand read stalls
- I5: Subtract to produce F11 using F9 and F5
        - Adder not available
- End state:
        - I2 executing, I3 stalled in Operand Read (RAW on F5), I4 stalled in Operand Read (RAW on F7), I5 cannot dispatch
        - Integer unit free, Multiplier unit reserved by I2, Adder unit reserved by I3, Divider unit reserved by I4
        - F5(I2), F7(I3) and F9(I4) pending. F3 ready. Other registers unchanged.

Cycle 12:
- I2: Multiply F5 using operands F3 and F7
        - Commits F5
- I3: Add F7 using operands F5 and F9
        - F9 available but F5 is pending
        - Operand read stalls
- I4: Divide F9 using operands F7 and F3
        - F3 is available but F7 is pending by I3
        - Operand read stalls
- I5: Subtract to produce F11 using F9 and F5
        - Adder not available
- End state:
        - I2 commits F5 and leaves, I3 stalled in Operand Read (RAW on F5), I4 stalled in Operand Read (RAW on F7), I5 cannot dispatch
        - Integer unit free, Multiplier unit free, Adder unit reserved by I3, Divider unit reserved by I4
        - F7(I3) and F9(I4) pending. F3 and F5 ready. Other registers unchanged.

Cycle 13:
- I3: Add F7 using operands F5 and F9
        - F5 and F9 is available
        - Successfully read operands
- I4: Divide F9 using operands F7 and F3
        - F3 is available but F7 is pending by I3
        - Operand read stalls
- I5: Subtract to produce F11 using F9 and F5
        - Adder not available
- End state:
        - I3 successfully read operands, I4 stalled in Operand Read (RAW on F7), I5 cannot dispatch
        - Integer unit free, Multiplier unit free, Adder unit reserved by I3, Divider unit reserved by I4
        - F7(I3) and F9(I4) pending. F3 and F5 ready. Other registers unchanged.

Cycle 14-17:
- I3: Add F7 using operands F5 and F9
        - Starts execution
        - Will execute for 4 cycles
        - Will complete execution by the end of cycle 17
- I4: Divide F9 using operands F7 and F3
        - F3 is available but F7 is pending by I3
        - Operand read stalls
- I5: Subtract to produce F11 using F9 and F5
        - Adder not available
- End state:
        - I3 executing, I4 stalled in Operand Read (RAW on F7), I5 cannot dispatch
        - Integer unit free, Multiplier unit free, Adder unit reserved by I3, Divider unit reserved by I4
        - F7(I3) and F9(I4) pending. F3 and F5 ready. Other registers unchanged.

Cycle 18:
- I3: Add F7 using operands F5 and F9
        - Commits F7
- I4: Divide F9 using operands F7 and F3
        - F3 is available but F7 is pending by I3
        - Operand read stalls
- I5: Subtract to produce F11 using F9 and F5
        - Adder not available
- End state:
        - I3 commits F7 and leaves, I4 stalled in Operand Read (RAW on F7), I5 cannot dispatch
        - Integer unit free, Multiplier unit free, Adder unit free, Divider unit reserved by I4
        - F9(I4) pending. F3, F5 and F7 ready. Other registers unchanged.

Cycle 19:
- I4: Divide F9 using operands F7 and F3
        - F3 and F7 both available
        - Successfully acquire operands
- I5: Subtract to produce F11 using F9 and F5
        - Adder available
        - Dispatches I5
- End state:
        - I4 successfully acquire operands, I5 dispatched
        - Integer unit free, Multiplier unit free, Adder reserved by I5, Divider unit reserved by I4
        - F9(I4), F11(I5) pending. F3, F5 and F7 ready. Other registers unchanged.

Cycle 20:
- I4: Divide F9 using operands F7 and F3
        - Starts execution
        - Will execute for 9 cycles
        - Will complete execution by 28
- I5: Subtract to produce F11 using F9 and F5
        - F5 is available but F9 is pending
        - Operand read stalls
- I6: Load F1 from the memory location addressed by R5
        - Integer unit available
        - Dispatches I6
- End state:
        - I4 starts execution, I5 stalled in Operand Read (RAW on F9), I6 dispatched
        - Integer reserved by I6, Multiplier unit free, Adder reserved by I5, Divider unit reserved by I4
        - F1(I6), F9(I4), F11(I5) pending. F3, F5 and F7 ready. Other registers unchanged.

Cycle 21:
- I4: Divide F9 using operands F7 and F3
        - Executing
- I5: Subtract to produce F11 using F9 and F5
        - F5 is available but F9 is pending
        - Operand read stalls
- I6: Load F1 from the memory location addressed by R5
        - R5 available
        - Successfully acquire operands
- I7: Multiply F3 using operands F1 and F11
        - Multiplier unit free
        - I7 dispatches
- End state:
        - I4 executing, I5 stalled in Operand Read (RAW on F9), I6 successfully acquire operands, I7 dispatched
        - Integer reserved by I6, Multiplier reserved by I7, Adder reserved by I5, Divider unit reserved by I4
        - F1(I6), F3(I7), F9(I4), F11(I5) pending. F5 and F7 ready. Other registers unchanged.

Cycle 22:
- I4: Divide F9 using operands F7 and F3
        - Executing
- I5: Subtract to produce F11 using F9 and F5
        - F5 is available but F9 is pending
        - Operand read stalls
- I6: Load F1 from the memory location addressed by R5
        - Starts execution
        - completes execution by the end of cycle
- I7: Multiply F3 using operands F1 and F11
        - F1 and F11 both are not available.
        - Operand read stalls
- I8: Add F5 using operands F3 and F7
        - Adder not available
        - Cannot dispatch
- End state:
        - I4 executing, I5 stalled in Operand Read (RAW on F9), I6 executing, I7 stalled in Operand Read (RAW on F1 and F11), I8 not dispatched
        - Integer reserved by I6, Multiplier reserved by I7, Adder reserved by I5, Divider unit reserved by I4
        - F1(I6), F3(I7), F9(I4), F11(I5) pending. F5 and F7 ready. Other registers unchanged.

Cycle 23:
- I4: Divide F9 using operands F7 and F3
        - Executing
- I5: Subtract to produce F11 using F9 and F5
        - F5 is available but F9 is pending
        - Operand read stalls
- I6: Load F1 from the memory location addressed by R5
        - Commits F1
- I7: Multiply F3 using operands F1 and F11
        - F1 is available but F11 is not available.
        - Operand read stalls
- I8: Add F5 using operands F3 and F7
        - Adder not available
        - Cannot dispatch
- End state:
        - I4 executing, I5 stalled in Operand Read (RAW on F9), I6 commit F1 and leaves, I7 stalled in Operand Read (RAW on F1 and F11), I8 not dispatched
        - Integer unit free, Multiplier reserved by I7, Adder reserved by I5, Divider unit reserved by I4
        - F3(I7), F9(I4), F11(I5) pending. F1, F5 and F7 ready. Other registers unchanged.

Cycle 24-28:
- I4: Divide F9 using operands F7 and F3
        - Executing
        - Completes execution in cycle 28
- I5: Subtract to produce F11 using F9 and F5
        - F5 is available but F9 is pending
        - Operand read stalls
- I7: Multiply F3 using operands F1 and F11
        - F1 is available but F11 is not available.
        - Operand read stalls
- I8: Add F5 using operands F3 and F7
        - Adder not available
        - Cannot dispatch
- End state:
        - I4 executing, I5 stalled in Operand Read (RAW on F9), I7 stalled in Operand Read (RAW on F1 and F11), I8 not dispatched
        - Integer unit free, Multiplier reserved by I7, Adder reserved by I5, Divider unit reserved by I4
        - F3(I7), F9(I4), F11(I5) pending. F1, F5 and F7 ready. Other registers unchanged.

Cycle 29:
- I4: Divide F9 using operands F7 and F3
        - Commits F9
- I5: Subtract to produce F11 using F9 and F5
        - F5 is available but F9 is pending
        - Operand read stalls
- I7: Multiply F3 using operands F1 and F11
        - F1 is available but F11 is not available.
        - Operand read stalls
- I8: Add F5 using operands F3 and F7
        - Adder not available
        - Cannot dispatch
- End state:
        - I4 commits F9 and leaves, I5 stalled in Operand Read (RAW on F9), I7 stalled in Operand Read (RAW on F1 and F11), I8 not dispatched
        - Integer unit free, Multiplier reserved by I7, Adder reserved by I5, Divider unit free
        - F3(I7), F11(I5) pending. F1, F5, F7 and F9 ready. Other registers unchanged.

Cycle 30:
- I5: Subtract to produce F11 using F9 and F5
        - F5 and F9 both are available
        - Successfully acquire operands
- I7: Multiply F3 using operands F1 and F11
        - F1 is available but F11 is not available.
        - Operand read stalls
- I8: Add F5 using operands F3 and F7
        - Adder not available
        - Cannot dispatch
- End state:
        - I5 successfully acquire operands, I7 stalled in Operand Read (RAW on F1 and F11), I8 not dispatched
        - Integer unit free, Multiplier reserved by I7, Adder reserved by I5, Divider unit free
        - F3(I7), F11(I5) pending. F1, F5, F7 and F9 ready. Other registers unchanged.

Cycle 31-34:
- I5: Subtract to produce F11 using F9 and F5
        - Execution starts
        - Will be executing for 4 cycles
        - Will be completing execution by the end of cycle 34
- I7: Multiply F3 using operands F1 and F11
        - F1 is available but F11 is not available.
        - Operand read stalls
- I8: Add F5 using operands F3 and F7
        - Adder not available
        - Cannot dispatch
- End state:
        - I5 starts execution, I7 stalled in Operand Read (RAW on F1 and F11), I8 not dispatched
        - Integer unit free, Multiplier reserved by I7, Adder reserved by I5, Divider unit free
        - F3(I7), F11(I5) pending. F1, F5, F7 and F9 ready. Other registers unchanged.

Cycle 35:
- I5: Subtract to produce F11 using F9 and F5
        - Commit F11
- I7: Multiply F3 using operands F1 and F11
        - F1 is available but F11 is not available.
        - Operand read stalls
- I8: Add F5 using operands F3 and F7
        - Adder not available
        - Cannot dispatch
- End state:
        - I5 commit F11 and leaves, I7 stalled in Operand Read (RAW on F1 and F11), I8 not dispatched
        - Integer unit free, Multiplier reserved by I7, Adder unit free, Divider unit free
        - F3(I7) pending. F1, F5, F7, F9 and F11 ready. Other registers unchanged.

Cycle 36:
- I7: Multiply F3 using operands F1 and F11
        - F1 and F11 both are available.
        - Successfully acquire operands
- I8: Add F5 using operands F3 and F7
        - Adder is available
        - I8 dispatches
- End state:
        - I7 successfully acquire operands, I8 dispatched
        - Integer unit free, Multiplier reserved by I7, Adder reserved by I8, Divider unit free
        - F3(I7), F5(I8) pending. F1, F7 and F9 ready. Other registers unchanged.

Cycle 37:
- I7: Multiply F3 using operands F1 and F11
        - Starts execution
        - Will be executing for 6 cycles
        - Will complete execution by cycle 42
- I8: Add F5 using operands F3 and F7
        - F7 is available but F3 is pending
        - Operand read stalls
- I9: Divide F7 using operands F5 and F1
        - Divider unit free
        - I9 dispatches
- End state:
        - I7 starts execution, I8 stalled in Operand Read (RAW on F3), I9 dispatches
        - Integer unit free, Multiplier reserved by I7, Adder reserved by I8, Divider reserved by I9
        - F3(I7), F5(I8), F7(I9) pending. F1 and F9 ready. Other registers unchanged.

Cycle 38:
- I7: Multiply F3 using operands F1 and F11
        - Executing
- I8: Add F5 using operands F3 and F7
        - F7 is available but F3 is pending
        - Operand read stalls
- I9: Divide F7 using operands F5 and F1
        - F1 is available but F5 is pending
        - Operand read stalls
- I10: Subtract to produce F9 using F7 and F3
        - Adder not available
        - Cannot dispatch I10
- End state:
        - I7 executing, I8 stalled in Operand Read (RAW on F3), I9 stalled in Operand Read (RAW on F5), I0 cannot dispatch
        - Integer unit free, Multiplier reserved by I7, Adder reserved by I8, Divider reserved by I9
        - F3(I7), F5(I8), F7(I9) pending. F1 and F9 ready. Other registers unchanged.

Cycle 39-42:
- I7: Multiply F3 using operands F1 and F11
        - Executing
        - Completes execution by the end of cycle 42
- I8: Add F5 using operands F3 and F7
        - F7 is available but F3 is pending
        - Operand read stalls
- I9: Divide F7 using operands F5 and F1
        - F1 is available but F5 is pending
        - Operand read stalls
- I10: Subtract to produce F9 using F7 and F3
        - Adder not available
        - Cannot dispatch I10
- End state:
        - I7 completes execution by the end of cycle 42, I8 stalled in Operand Read (RAW on F3), I9 stalled in Operand Read (RAW on F5), I0 cannot dispatch
        - Integer unit free, Multiplier reserved by I7, Adder reserved by I8, Divider reserved by I9
        - F3(I7), F5(I8), F7(I9) pending. F1 and F9 ready. Other registers unchanged.

Cycle 43:
- I7: Multiply F3 using operands F1 and F11
        - Commits F3
- I8: Add F5 using operands F3 and F7
        - F7 is available but F3 is pending
        - Operand read stalls
- I9: Divide F7 using operands F5 and F1
        - F1 is available but F5 is pending
        - Operand read stalls
- I10: Subtract to produce F9 using F7 and F3
        - Adder not available
        - Cannot dispatch I10
- End state:
        - I7 commits F3 and leaves, I8 stalled in Operand Read (RAW on F3), I9 stalled in Operand Read (RAW on F5), I0 cannot dispatch
        - Integer unit free, Multiplier unit free, Adder reserved by I8, Divider reserved by I9
        - F5(I8), F7(I9) pending. F1, F3 and F9 ready. Other registers unchanged.

Cycle 44:
- I8: Add F5 using operands F3 and F7
        - F3 and F7 are available
        - Successfully acquire operands
- I9: Divide F7 using operands F5 and F1
        - F1 is available but F5 is pending
        - Operand read stalls
- I10: Subtract to produce F9 using F7 and F3
        - Adder not available
        - Cannot dispatch I10
- End state:
        - I8 successfully acquire operands, I9 stalled in Operand Read (RAW on F5), I10 cannot dispatch
        - Integer unit free, Multiplier unit free, Adder reserved by I8, Divider reserved by I9
        - F5(I8), F7(I9) pending. F1, F3 and F9 ready. Other registers unchanged.

Cycle 45-48:
- I8: Add F5 using operands F3 and F7
        - Starts execution
        - Will execute for 4 cycles
        - Will complete execution by the end of cycle 48
- I9: Divide F7 using operands F5 and F1
        - F1 is available but F5 is pending
        - Operand read stalls
- I10: Subtract to produce F9 using F7 and F3
        - Adder not available
        - Cannot dispatch I10
- End state:
        - I8 executing, I9 stalled in Operand Read (RAW on F5), I10 cannot dispatch
        - Integer unit free, Multiplier unit free, Adder reserved by I8, Divider reserved by I9
        - F5(I8), F7(I9) pending. F1, F3 and F9 ready. Other registers unchanged.

Cycle 49:
- I8: Add F5 using operands F3 and F7
        - Commits F5
- I9: Divide F7 using operands F5 and F1
        - F1 is available but F5 is pending
        - Operand read stalls
- I10: Subtract to produce F9 using F7 and F3
        - Adder not available
        - Cannot dispatch I10
- End state:
        - I8 commits F5 and leaves, I9 stalled in Operand Read (RAW on F5), I10 cannot dispatch
        - Integer unit free, Multiplier unit free, Adder unit free, Divider reserved by I9
        - F7(I9) pending. F1, F3, F5 and F9 ready. Other registers unchanged.

Cycle 50:
- I9: Divide F7 using operands F5 and F1
        - F1 and F5 both are available
        - Successfully acquire operands
- I10: Subtract to produce F9 using F7 and F3
        - Adder is available
        - I10 dispatches
- End state:
        - I9 successfully acquire operands, I0 dispatched
        - Integer unit free, Multiplier unit free, Adder unit reserved by I10, Divider reserved by I9
        - F7(I9), F9(I10) pending. F1, F3, F5 ready. Other registers unchanged.

Cycle 51-59:
- I9: Divide F7 using operands F5 and F1
        - Starts execution
        - Will execute for 9 cycles
        - Will complete execution by the end of cycle 59
- I10: Subtract to produce F9 using F7 and F3
        - F3 is available but F7 is pending
        - Operand read stalls
- End state:
        - I9 starts execution, I0 stalled in Operand Read (RAW on F7)
        - Integer unit free, Multiplier unit free, Adder unit reserved by I10, Divider reserved by I9
        - F7(I9), F9(I10) pending. F1, F3, F5 ready. Other registers unchanged.

Cycle 60:
- I9: Divide F7 using operands F5 and F1
        - Commits F7
- I10: Subtract to produce F9 using F7 and F3
        - F3 is available but F7 is pending
        - Operand read stalls
- End state:
        - I9 commit F7 and leaves, I0 stalled in Operand Read (RAW on F7)
        - Integer unit free, Multiplier unit free, Adder unit reserved by I10, Divider unit free
        - F9(I10) pending. F1, F3, F5, F7 ready. Other registers unchanged.

Cycle 61:
- I10: Subtract to produce F9 using F7 and F3
        - F3 and F7 both are available
        - Successfully acquire operands
- End state:
        - I0 successfully acquire operands
        - Integer unit free, Multiplier unit free, Adder unit reserved by I10, Divider unit free
        - F9(I10) pending. F1, F3, F5, F7 ready. Other registers unchanged.

Cycle 62-65:
- I10: Subtract to produce F9 using F7 and F3
        - Starts execution
        - Will be executing for 4 cycles
        - Will complete execution by cycle 65
- End state:
        - I0 executing
        - Integer unit free, Multiplier unit free, Adder unit reserved by I10, Divider unit free
        - F9(I10) pending. F1, F3, F5, F7 ready. Other registers unchanged.

Cycle 66:
- I10: Subtract to produce F9 using F7 and F3
        - Commit F9
- End state:
        - I0 commits F9 and leavs
        - Integer unit free, Multiplier unit free, Adder unit free, Divider unit free
        - F1, F3, F5, F7 and F9 ready. Other registers unchanged.

Final answers
1. The cycle at which the final instruction finishes committing its result
Cycle 66

2. The total of all dispatch cycle numbers
1 + 2 + 3 + 4 + 19 + 20 + 21 + 36 + 37 + 50 = 193

3. The total of all result-commit cycle numbers
339

4. The cumulative number of stall cycles caused by WAW conflicts
0

5. The commit cycle of instruction I6
23

6. The earliest cycle at which the integer execution unit becomes permanently idle
24

7. The maximum number of instructions simultaneously present in the system (issued but not yet committed)
5

8. The instruction that experiences the longest continuous stall period
I4

9. The total number of cycles during which no instruction is able to dispatch
56

10. The average execution-unit utilization across all functional units over the entire timeline
| Unit        | Busy cycles |
| ----------- | ----------- |
| Integer     | 2           |
| FP Add/Sub  | 16                    |
| FP Multiply | 12          |
| FP Divide   | 18          |

Total possible = 4 × 66 = 264
Utilization = 48/264 = 18.2%


Prompt:
You are given multiple test cases. In each test case, you are given an array `p` of length `n` describing a permutation of `{1,2,…,n}` with missing elements: some `p[i] = -1`, and all non-`-1` values in `p` are **distinct**.

A permutation `b` of length `n` is called a **riffle shuffle** of the sorted permutation `[1,2,…,n]` if there exists an integer `k` with `1 ≤ k < n` such that:
- the sequence `1,2,…,k` appears as a subsequence of `b`, and
- the sequence `k+1,k+2,…,n` appears as a subsequence of `b`.

Your task: for each test case, count the number of distinct ways to replace every `-1` in `p` with the missing integers (so that the result is a full permutation of `1..n`) such that the resulting permutation is a riffle shuffle of `[1..n]`.

Output the answer modulo `998244353`.

### Input format
- First line: integer `t` (number of test cases).
For each test case:
- Line 1: integer `n`.
- Line 2: `n` integers `p[1..n]` (each is `-1` or in `[1..n]`), with all non-`-1` distinct.

### Output format
For each test case, output one integer: the number of valid completions modulo `998244353`.

For a known value v at position i with v<=k, the answer states the exact prefix constraint B[i]=v (i.e., exactly v BLUE positions among indices 1..i).

For a known value v at position i with v>k, the answer states the exact prefix constraint B[i]=i-(v-k).

The answer explicitly states that if two constraints assign different required values to the same prefix length j, then that split k contributes 0 (is discarded).

The answer computes the number of valid colorings in a feasible block as exactly the binomial coefficient C(len, need).

The answer includes an explicit correction for the sorted completion case (when compatible with the partial p) to avoid overcounting across k (e.g., subtracting n-2 modulo 998244353).

The answer reformulates the problem for a fixed split k by coloring values v<=k as BLUE and values v>k as RED.

For a known value v at position i with v<=k, the answer states the exact constraint B[i-1]=v-1 (forcing position i itself to be BLUE).

The answer states the feasibility condition between consecutive constrained prefixes: for a block with len=jb-ja and need=Bb-Ba, it must hold that 0<=need<=len.

The answer includes an aggregation step that sums count(k) over all k from 1 to n-1 to form the total.

For a known value v at position i with v>k, the answer states the exact constraint B[i-1]=B[i] (forcing position i itself to be RED).

Let MOD = 998244353. Precompute factorials `fact[i]` and inverse factorials `invfact[i]` up to the maximum `n` across all test cases, so `C(n,r) = fact[n]*invfact[r]*invfact[n-r] mod MOD`.

Key uniqueness fact (split-point identification):
- For any riffle shuffle permutation `q` that is **not** fully sorted, the split value `k` is uniquely determined as the smallest integer such that `k` appears after `k+1` in `q`.
- The only riffle shuffle permutation where no such `k` exists is the sorted permutation `[1,2,…,n]`, which is valid for every `k` and therefore must be handled to avoid overcounting.

Fix a split value `k` (1 ≤ k < n). Color values `1..k` as BLUE and `k+1..n` as RED. A completion is valid for this `k` iff the BLUE values appear in increasing order and the RED values appear in increasing order; once the BLUE/RED pattern of positions is fixed, the final permutation is uniquely determined by scanning left-to-right and assigning the next unused BLUE number or RED number.

Prefix-count reformulation (explicit constraints):
Define `B[j]` = number of BLUE positions among the first `j` positions (prefix length `j`), for `j = 0..n`.
Then necessarily:
- `B[0] = 0`
- `B[n] = k`
- For every position `i` with known value `v = p[i] != -1`:

  1) If `v ≤ k` (BLUE value), then:
     - `B[i] = v`  (exactly `v` blues in positions `1..i`)
     - `B[i-1] = v-1` (position `i` itself must be BLUE)

  2) If `v > k` (RED value), then the number of reds in `1..i` must be `v-k`, so:
     - `B[i] = i - (v-k)`  (since blues = i − reds)
     - `B[i-1] = B[i]` (position `i` itself must be RED)

If any prefix length `j` gets two different required values, this `k` is impossible.

Counting for a fixed k using binomials:
Let `req[j]` store the required value of `B[j]` (or “unset”). Initialize:
- `req[0]=0`, `req[n]=k`, and add the above constraints from every known `p[i]`.

Collect all prefix indices `j` where `req[j]` is set, sort them increasingly: `j0 < j1 < ... < jm`.
For each consecutive pair `(ja, Ba=req[ja])` and `(jb, Bb=req[jb])`:
- `len = jb - ja` (number of positions in this block)
- `need = Bb - Ba` (number of BLUE positions in that block)
Feasibility condition (must hold):
- `0 ≤ need ≤ len`
If not, this `k` is impossible.
Otherwise, the number of ways to choose which positions in this block are BLUE is exactly:
- `C(len, need)`
Multiply these contributions over all blocks to get `count(k)`.

Sorted permutation special case:
Check whether the fully sorted permutation `[1..n]` is compatible with the partial input: i.e., for every `i`, if `p[i] != -1` then `p[i] == i`. If compatible, there is exactly one “sorted completion”; call this `sorted_ok = 1`, else `0`.

Final aggregation:
Compute `S = sum_{k=1..n-1} count(k) mod MOD`.
If `sorted_ok = 1`, the sorted completion was counted once for every `k` (i.e., `n-1` times), but must be counted once total, so subtract `(n-2)`:
- `answer = (S - (n-2)) mod MOD`
Else:
- `answer = S`
Output `answer` modulo MOD (normalized to [0, MOD)).

(Implementation note: `C(len, need)` is computed in O(1) using factorials and inverse factorials.)


Prompt:
You are given an integer sequence `a = [a1, a2, …, an]`, where each value represents the height of a structure placed in a line.

### Visibility rule (record highs)
Define two collections derived from a sequence `x`:

- **Left-visible heights** `L(x)`: scan `x` from left to right and write down a height whenever it is **strictly greater** than every height seen earlier in the scan.
- **Right-visible heights** `R(x)`: scan `x` from right to left and write down a height whenever it is **strictly greater** than every height seen earlier in that right-to-left scan (equivalently: strictly greater than all elements to its right).

These collections are treated as **sets of height values** (duplicates are irrelevant).

#### Example
For `x = [2, 4, 4, 1, 5, 3]`:
- `L(x) = {2, 4, 5}`
- `R(x) = {3, 5}`

### What you must count
A **subsequence** `a'` is obtained by choosing a subset of indices `1 ≤ i1 < i2 < … < ik ≤ n` and taking:
`a' = [ai1, ai2, …, aik]`.

Count how many subsequences `a'` satisfy **both** of the following:
- `L(a') = L(a)`
- `R(a') = R(a)`

Two subsequences are considered different if they are formed from different index sets, even if their values are identical.

Because the answer can be large, output it modulo:

`998244353`.

---

## Input
The first line contains an integer `t` — the number of test cases.

For each test case:
- The first line contains an integer `n` (`1 ≤ n ≤ 5000`)
- The second line contains `n` integers `a1, a2, …, an` (`1 ≤ ai ≤ 10^9`)

Additional constraint:
- The sum of `n` over all test cases does not exceed `5000`.

---

## Output
For each test case, print a single integer: the number of subsequences `a'` such that
`L(a') = L(a)` and `R(a') = R(a)`, modulo `998244353`.

Does the answer define L(a) as the record-high values when scanning left to right using a strict 'greater than all previous' rule?

Does the answer state that when processing a non-maximum element x with x <= the last visible value of the current prefix, the DP value for that prefix is multiplied by 2 (take/skip)?

Does the answer specify dpL[i] for positions with a[i] = m as the number of subsequences that match L(a) excluding its last element (expressed as dpL[i] = c[k-1] or equivalent)?

Does the answer give the final sum formula over i <= j with a[i] = a[j] = m as dpL[i] * dpR[j] * 2^(j-i-1) modulo 998244353?

Does the answer state that the final result is computed modulo 998244353?

Does the answer state that when processing a non-maximum element x that equals the next required visible value L[p], the transition adds c[p] into c[p+1]?

Does the answer state that the computed right-DP results are converted back to the original indexing?

Does the answer define m as max(a) (the maximum element of the sequence)?

Does the answer state the DP initialization c[0] = 1 (with other c values initially 0)?

Does the answer define R(a) as the record-high values when scanning right to left using a strict 'greater than all after' rule?

### Overview
We need to count subsequences `a'` whose **left-visible set** and **right-visible set** match those of the original sequence `a`.

Let:
- `L(a)` be the (strictly increasing) sequence of record highs seen from the left.
- `R(a)` be the (strictly increasing, in right-to-left scan order) record highs seen from the right.
- `m = max(a)`. The value `m` must appear in any valid subsequence and is visible from both sides.

The main idea is to **fix the first and last occurrence of `m` inside the subsequence**, then multiply independent counts for:
1) building the required left view up to the first `m`,
2) building the required right view from the last `m`,
3) freely choosing elements strictly between those two maxima.

All arithmetic is done modulo `998244353`.

---

### Step 1: Compute `L(a)` and `R(a)`
Scan:
- From left to right, append `a[i]` to `L` if it is strictly greater than all previous.
- From right to left, append `a[i]` to `R` if it is strictly greater than all to its right.

Let `L` have length `k`. Note that `L[k-1] = m`.

---

### Step 2: Count valid prefixes ending at each maximum (`dpL`)
For each position `i` with `a[i] = m`, define:

- `dpL[i]` = number of subsequences formed using indices `≤ i`
  such that their left-visible sequence equals `L` **without its last element** (i.e., the first `k-1` record highs),
  and additionally **no `m` is chosen before position `i`**.

Then, if we append `a[i]=m` as the first maximum, the left view becomes exactly `L(a)`.

#### How to compute `dpL`
Maintain an array `c[0..k-1]` where:
- `c[p]` = number of subsequences whose left-visible sequence equals the first `p` elements of `L`
  (so `p` ranges from `0` to `k-1`, since we do *not* want to place `m` yet).

Initialization:
- `c[0] = 1`, all other `c[p] = 0`.

Process elements from left to right. For each `x = a[i]`:
- If `x == m`:
  - set `dpL[i] = c[k-1]` (these are exactly the subsequences that already match the first `k-1` visible heights),
  - and **do not** allow selecting `x` into `c` (we skip it in the DP so that future maxima can still be “first maximum” candidates).

- Otherwise, update `c` to reflect the choice “take `x`” or “skip `x`”, under the constraint that the left-visible sequence must remain a prefix of `L`:

For each prefix length `p`:
- Let `last = (p == 0 ? -∞ : L[p-1])`.
- If `x <= last`, then taking or skipping `x` does not change the visible record highs:
  - `c[p] := 2 * c[p]`.
- Else (`x > last`), the only way to take `x` without breaking the required left-visible sequence is if it equals the *next* needed visible height:
  - if `p < k-1` and `x == L[p]`, then we can extend:
    - `c[p+1] += c[p]` (take `x`),
    - and also keep `c[p]` as-is (skip `x`).
  - if `x > last` but `x != L[p]`, we cannot take `x` (only skipping is allowed), so `c[p]` stays unchanged.

This DP runs in `O(n * k)`, and `k ≤ n`, so `O(n^2)`.

---

### Step 3: Count valid suffixes starting at each maximum (`dpR`)
Similarly define for each position `j` with `a[j] = m`:

- `dpR[j]` = number of subsequences using indices `≥ j`
  such that their right-visible sequence equals `R(a)` without its last element,
  and **no `m` is chosen after position `j`**.

Compute `dpR` by running the same DP logic on the reversed sequence (and using the record highs of `R(a)` in the appropriate order), then map results back to original indices.

---

### Step 4: Combine by fixing first/last maximum
Fix positions `i ≤ j` with `a[i] = a[j] = m`, where:
- `i` is the first chosen maximum in the subsequence,
- `j` is the last chosen maximum in the subsequence.

Elements strictly between `i` and `j` can be chosen arbitrarily: they are “sandwiched” between maxima and cannot affect visibility from either side.

Let `x = j - i - 1` be the number of elements between them. Then the number of choices is `2^x`.

Contribution of pair `(i, j)`:
- `dpL[i] * dpR[j] * 2^(j-i-1)`.

So the final answer is:
\[
\text{ans} = \sum_{\substack{i \le j \\ a[i]=a[j]=m}} dpL[i] \cdot dpR[j] \cdot 2^{(j-i-1)} \pmod{998244353}.
\]

Precompute powers of two up to `n` modulo `998244353`.

---

### Complexity
- Computing `L(a)` and `R(a)`: `O(n)`
- Computing `dpL` and `dpR`: `O(n^2)`
- Summing over maximum pairs: `O(n^2)` in the worst case

Total per test file fits the constraints because `∑n ≤ 5000`.

---


Prompt:
# Operating Systems / Storage (RAID): Deterministic Layout, Parity Update, Rebuild, and Reliability

## Global constants (used everywhere)

- Disks are named: `D0, D1, D2, D3, D4` (5 disks total).
- Sector size is irrelevant except for parity; for this task **each sector stores exactly 1 byte**.
- A stripe unit (chunk) is **4 sectors**.
- All XOR parity is byte-wise XOR (8-bit).

### Common stripe indexing
Let logical address be in **sectors**:
- RAID5 has **4 data disks per stripe** ⇒ logical stripe width:
  \[
  W_{5}=4\cdot 4=16 \text{ logical sectors per stripe}
  \]
- Stripe number:
  \[
  s = \left\lfloor \frac{\text{LBA}}{16}\right\rfloor
  \]
- Position within stripe:
  \[
  p = \text{LBA}\bmod 16
  \]
- Data-chunk index within stripe:
  \[
  j = \left\lfloor \frac{p}{4}\right\rfloor \in \{0,1,2,3\}
  \]
- Offset within chunk:
  \[
  o = p \bmod 4 \in \{0,1,2,3\}
  \]
- Physical sector number on a disk for stripe `s` at offset `o` is:
  \[
  \text{physSector} = 4s + o
  \]

---

## Part A — RAID 5 (distributed parity, **layout is fixed and must be used**)

### RAID5 layout rule (left-asymmetric, deterministic)
For stripe number `s`:
- Parity disk:
  \[
  pDisk = s \bmod 5
  \]
- The 4 data chunks `j=0..3` are placed on disks:
  \[
  dDisk(j) = (pDisk + 1 + j) \bmod 5
  \]
- For any stripe `s` and offset `o`, parity byte stored on `pDisk` at sector `4s+o` is:
  \[
  P[s,o] = D[s,0,o]\oplus D[s,1,o]\oplus D[s,2,o]\oplus D[s,3,o]
  \]
  where `D[s,j,o]` is the data byte for chunk `j` offset `o`.

### RAID5 read requests (must map to physical disk sector reads)
Compute the minimal set of physical read operations for each request, using this rule:

**Read-I/O grouping rule:** within the same disk, merge any reads that touch **contiguous sector ranges** into one `(disk,start,len)` tuple.

Output tuples must be sorted by disk number ascending and, within a disk, by `start` ascending.

#### RAID5 Read Request R1
Read `len=22` logical sectors starting at `LBA=14` (i.e., LBAs 14..35).

#### RAID5 Read Request R2
Read `len=18` logical sectors starting at `LBA=7` (i.e., LBAs 7..24).

---

## Part B — RAID5 write W (parity update) and I/O counts

### Deterministic write W
Write `len=11` logical sectors starting at `LBA=23` (LBAs 23..33).
The new sector bytes (in order of increasing LBA) are:

`A0 A1 A2 A3 A4 A5 A6 A7 A8 A9 AA` (hex bytes, uppercase, no `0x`).

### RAID5 initial on-disk bytes (only stripes used by W are provided)

Stripe `s=1` (so sectors `4..7` on each disk for offsets `o=0..3`):
- Parity disk is `D1`. Data disks are `D2 (j=0), D3 (j=1), D4 (j=2), D0 (j=3)`.

Given **data bytes** (offsets `o=0,1,2,3`):
- `D2: 10 11 12 13`
- `D3: 20 21 22 23`
- `D4: 30 31 32 33`
- `D0: 40 41 42 43`

Stripe `s=2` (sectors `8..11` on each disk):
- Parity disk is `D2`. Data disks are `D3 (j=0), D4 (j=1), D0 (j=2), D1 (j=3)`.

Given **data bytes**:
- `D3: 55 56 57 58`
- `D4: 65 66 67 68`
- `D0: 75 76 77 78`
- `D1: 05 06 07 08`

### RAID5 write-parity update rule (must follow exactly)
Write uses **read-modify-write** with these exact rules:

For each stripe `s` touched by the write:
1. Read the **entire parity chunk** for that stripe: 4 contiguous sectors on `pDisk` → counts as **1 read I/O**.
2. For each data disk that has any sectors overwritten in that stripe, read the overwritten sectors on that disk; within that disk, merge contiguous reads → counts as read I/O(s).
3. Write the new data sectors (merged per disk into contiguous runs) → counts as write I/O(s).
4. Write back the **entire parity chunk** for that stripe (4 contiguous sectors) → counts as **1 write I/O**.

**Important:** Reads/writes on the same disk may be merged into a single I/O if they are contiguous *even if they belong to different stripes* (because physical sectors are contiguous). Use the same merge rule as in Part A.

### What you must compute for RAID5 write W
- `WRITE_W_IO_RAID5`: total read I/Os and total write I/Os for the entire write W.
- `RAID5_PARITY_STRIPE1`: the 4 parity bytes on `D1` sectors `4..7` after W completes, as comma-separated hex bytes.
- `RAID5_PARITY_STRIPE2`: the 4 parity bytes on `D2` sectors `8..11` after W completes.

---

## Part C — RAID 4 (dedicated parity) on the same 5 disks

RAID4 uses:
- Data disks: `D0..D3`
- Dedicated parity disk: `D4`

Chunk size is still 4 sectors, and logical stripe width is also \(16\) sectors.

### RAID4 layout rule (deterministic)
For stripe `s`:
- Chunk `j` is stored on disk `Dj` for `j=0..3` (so `j=0→D0`, `j=1→D1`, etc.).
- Parity is stored on `D4`, and for each offset `o`:
  \[
  P4[s,o] = D0[s,o]\oplus D1[s,o]\oplus D2[s,o]\oplus D3[s,o]
  \]
- Physical sector on any disk for stripe `s` offset `o` is `4s+o`.

### RAID4 initial bytes (only stripes used by W are provided)

Stripe `s=1` (sectors `4..7`):
- `D0: 01 02 03 04`
- `D1: 05 06 07 08`
- `D2: 09 0A 0B 0C`
- `D3: 0D 0E 0F 10`

Stripe `s=2` (sectors `8..11`):
- `D0: 21 22 23 24`
- `D1: 25 26 27 28`
- `D2: 29 2A 2B 2C`
- `D3: 2D 2E 2F 30`

(Parity on `D4` is implied by XOR; you may compute it.)

### RAID4 write behavior for W
Apply the same **read-modify-write I/O rules** as RAID5, except:
- parity disk is always `D4`.

### What you must compute for RAID4 under the same W
- `WRITE_W_IO_RAID4`: total read I/Os and total write I/Os.
- `RAID4_PARITY`: after W completes, output:
  - stripe1 parity bytes on `D4` sectors `4..7`
  - stripe2 parity bytes on `D4` sectors `8..11`

---

## Part D — RAID5 rebuild check (coupled to your parity result)
After RAID5 write W completes, assume **disk D2 fails** completely.

Compute the reconstructed byte that should be stored at:
- stripe `s=1`, chunk `j=0`, offset `o=1`

Output that reconstructed byte as **two hex digits**.

---

## Part E — Reliability (numeric, deterministic rounding)

Let per-disk failure probability over the period be:
\[
p = 0.03
\]
Assume independent disk failures.

Compute data-loss probability (rounded to **exactly 10 digits after the decimal point**):

1. `P_LOSS_RAID0_5DISKS_p0.03`: RAID0 on 5 disks (loss if **any** disk fails).
2. `P_LOSS_RAID5_5DISKS_p0.03`: RAID5 on 5 disks (loss if **2 or more** disks fail).
3. `P_LOSS_RAID1_2DISKS_p0.03`: RAID1 mirrored pair (2 disks; loss if **both** fail).

---

## Required Output Format

Provide your reasoning, then output a final answer block containing **exactly 11 lines** in this exact order:

1. `RAID5_READ_REQ1: ...`
2. `RAID5_READ_REQ2: ...`
3. `WRITE_W_IO_RAID5: reads=N writes=N`
4. `WRITE_W_IO_RAID4: reads=N writes=N`
5. `RAID5_PARITY_STRIPE1: D1[4..7]=..,..,..,..`
6. `RAID5_PARITY_STRIPE2: D2[8..11]=..,..,..,..`
7. `RAID4_PARITY: stripe1 D4[4..7]=..,..,..,.. | stripe2 D4[8..11]=..,..,..,..`
8. `REBUILD_RAID5_D2_s1_j0_o1: ..`
9. `P_LOSS_RAID0_5DISKS_p0.03: 0.xxxxxxxxxx`
10. `P_LOSS_RAID5_5DISKS_p0.03: 0.xxxxxxxxxx`
11. `P_LOSS_RAID1_2DISKS_p0.03: 0.xxxxxxxxxx`

Line 3 is exactly `WRITE_W_IO_RAID5: reads=5 writes=5`.

Line 4 is exactly `WRITE_W_IO_RAID4: reads=5 writes=5`.

Line 5 is exactly `RAID5_PARITY_STRIPE1: D1[4..7]=34,34,34,BF`.

Line 6 is exactly `RAID5_PARITY_STRIPE2: D2[8..11]=BC,BC,40,40`.

Line 8 is exactly `REBUILD_RAID5_D2_s1_j0_o1: 11`.

Line 9 is exactly `P_LOSS_RAID0_5DISKS_p0.03: 0.1412659743`.

Line 10 is exactly `P_LOSS_RAID5_5DISKS_p0.03: 0.0084720528`.

Line 11 is exactly `P_LOSS_RAID1_2DISKS_p0.03: 0.0009000000`.

Does the final answer block contain exactly 11 lines in the specified order and format, regardless of any preceding reasoning?


Machine Learning & Artificial Intelligence

Prompt:
Output exactly one decimal number with exactly four digits after the decimal point.

Fixed data ($m=2$):
- $x^{(1)}=[1,0]^T,\ y^{(1)}=1$
- $x^{(2)}=[0,1]^T,\ y^{(2)}=0$

Fixed model:
- $b=0,\ w=[2,-2]^T$
- $\sigma(z)=\frac{1}{1+e^{-z}}$
- $z^{(i)}=w^T x^{(i)},\ \hat y^{(i)}=\sigma(z^{(i)})$

For any $\lambda>0$ define
$$
J_{\text{base},\lambda}(w)=\frac{1}{2}\sum_{i=1}^{2}\Big[-y^{(i)}\ln(\hat y^{(i)})-(1-y^{(i)})\ln(1-\hat y^{(i)})\Big]+\frac{\lambda}{2}\|w\|_2^2.
$$
Let $J_{0.1}=J_{\text{base},0.1}(w)$ and $J_{1.0}=J_{\text{base},1.0}(w)$.

Conventions: $\lfloor\cdot\rfloor$ is floor. For integers, $a\bmod M$ is the unique integer in $\{0,1,\dots,M-1\}$ congruent to $a$ modulo $M$. For nonnegative integers, $x>>k=\lfloor x/2^k\rfloor$ and $x\ \text{xor}\ y$ is bitwise XOR.

Set $\epsilon=10^{-12}$.

Define
$$
t=\frac{1}{2}\sum_{i=1}^{2}\Big[-y^{(i)}\ln(\hat y^{(i)})-(1-y^{(i)})\ln(1-\hat y^{(i)})\Big],
\quad
s=\hat y^{(1)}(1-\hat y^{(1)}),
\quad
\mu=0.1+\frac{s}{2}.
$$
Define the $2\times 2$ matrix $H=\mu I_2$ and $g=\left[\ 0.2-\frac{1}{2(1+e^2)},\ -0.2+\frac{1}{2(1+e^2)}\ \right]^T$.
Define $\gamma=1/\|w\|_2$, $R=\max_i\|x^{(i)}\|_2$, and $C=R\|w\|_2/\sqrt{2}$.
Define $\mathrm{tr}(H)=2\mu$ and $\det(H)=\mu^2$.

Define $\Delta=0$ and $\Omega=0$.

Define $w_\infty^\star=0$ and $L_\infty=\|w_\infty^\star\|_2^2$.
Define $\sigma^2>0$ by $1/\sigma^2=0.1$.

Define bits $b_0,\dots,b_{23}\in\{0,1\}$:
- $b_0=1$ iff $|\Delta|\le\epsilon$
- $b_1=1$ iff $\Omega\le\epsilon$
- $b_2=1$ iff $\mu\ge 0$
- $b_3=1$ iff $\|g\|_2>0$
- $b_4=1$ iff $|\gamma-1/\sqrt{8}|\le\epsilon$
- $b_5=1$ iff $J_{0.1}<J_{1.0}$
- $b_6=1$ iff $L_\infty\le\epsilon$
- $b_7=1$ iff $|0.1-1/\sigma^2|\le\epsilon$
- $b_8=1$ iff $\lfloor 10C\rfloor\bmod 2=1$
- $b_9=1$ iff $\lfloor 1000t\rfloor\bmod 2=1$
- $b_{10}=1$ iff $\mathrm{tr}(H)>0.304$
- $b_{11}=1$ iff $\det(H)\ge 0.0225$
- $b_{12}=1$ iff $\lfloor 10^6(\hat y^{(1)}-0.88)\rfloor\bmod 2=1$
- $b_{13}=1$ iff $\lfloor 10^6(\mu-0.15)\rfloor\bmod 2=1$
- $b_{14}=1$ iff $(\lfloor 10^4 t\rfloor\bmod 3)=0$
- $b_{15}=1$ iff $(\lfloor 10^6\det(H)\rfloor\bmod 7)\in\{0,3,5\}$
- $b_{16}=1$ iff $(\lfloor 10^6(\hat y^{(2)}-0.11)\rfloor\bmod 2)=1$
- $b_{17}=1$ iff $\lfloor 10^{12}(\hat y^{(1)}+\hat y^{(2)}-1)\rfloor=0$
- $b_{18}=1$ iff $\lfloor 10^{12}(\mu^2-\det(H))\rfloor=0$
- $b_{19}=1$ iff $\lfloor 10^5(J_{0.1}-t)\rfloor=40000$
- $b_{20}=1$ iff $\lfloor 10^{12}(C-2)\rfloor=0$
- $b_{21}=1$ iff $(\lfloor 10^6(\hat y^{(1)}-0.5)\rfloor\bmod 3)=1$
- $b_{22}=1$ iff $(\lfloor 10^6(\hat y^{(2)}-0.1)\rfloor\bmod 5)\in\{2,4\}$
- $b_{23}=1$ iff $(\lfloor 10^6(t-0.12)\rfloor\bmod 7)\in\{1,2,4\}$

Pack
$$
P=\sum_{k=0}^{23} b_k 2^k.
$$
Define Gray and bit-reverse:
$$
G=P\ \text{xor}\ (P>>1),
\quad
\mathrm{rev}_{24}(x)=\sum_{j=0}^{23}\left(\left\lfloor\frac{x}{2^j}\right\rfloor\bmod 2\right)2^{23-j},
\quad
R=\mathrm{rev}_{24}(G).
$$

Define
$$
S_0=(1103515245\cdot R+12345)\bmod 2^{31},
\quad
S_1=S_0\ \text{xor}\ (S_0>>11)\ \text{xor}\ (S_0>>21).
$$
Let $M=1000000$ and set
$$
T_0=S_1\bmod M,\quad
T_1=(17389\cdot T_0+711)\bmod M,\quad
T_2=(88411\cdot T_1+31337)\bmod M,
$$
$$
W=(T_2\ \text{xor}\ (T_2>>3)\ \text{xor}\ (T_2>>9))\bmod M.
$$

Finally define
$$
J(w)=J_{0.1}+\frac{W-500000}{1000000}.
$$

Do not round intermediate quantities. Round only the final $J(w)$ to four decimal places and output it.

Format: The final answer matches regex ^[0-9]+\.[0-9]{4}?


Prompt:
## Edge-interval cactus counting

You have an undirected graph on `N` vertices with `M` edges in a fixed order `E1..EM`.

A connected component is **cactus-like** if every edge belongs to at most one simple cycle.
The graph is **dune-stable** if each connected component is cactus-like.

Count pairs `(L, R)` with `1 ≤ L ≤ R ≤ M` such that the subgraph using only edges
`EL, EL+1, …, ER` is dune-stable.

### Input
- Two integers `N` and `M`
- Then `M` lines, the endpoints of edge `Ei`

### Output
- One integer, the number of valid `(L, R)` pairs

Does the response define a dune-stable graph as one whose every connected component is cactus-like?

Does the response state that dune-stability is preserved under deleting edges?

Does the response state that for each fixed L, the number of valid intervals starting at L equals R - L + 1 after maximizing R?

Does the response describe the edge-as-vertex transformation by introducing a vertex mid(e) for each original edge e?

Does the response state the invariant that each simple cycle has exactly one non-tree edge excluded from the maintained spanning forest?

Does the response state that inserting an edge that closes a cycle is rejected when the tree path between its endpoints contains a used-marked edge?

Does the response state that the excluded non-tree edge for a newly formed cycle is chosen as the minimum-index edge on that cycle?

Does the response state that marking cycle edges is implemented by a lazy path update that sets used to 1 on a link-cut tree path?

Does the response state that deleting an excluded non-tree cycle edge triggers a path update that sets used to 0 on the path between its endpoints?

Does the response state that deleting a tree edge is handled by cutting it from the maintained spanning forest?

## Expert solution

**Terminology.** The prompt calls a graph **dune-stable** if every connected component is cactus-like (each edge lies on at most one simple cycle). I will refer to this same property as **desert** below: **desert ⇔ dune-stable**.

### Monotonicity (enables two pointers)
- If a graph is desert (dune-stable), **deleting edges** cannot create a violation, so it stays desert.
- If a graph is **not** desert, **adding edges** cannot fix the existence of an edge that already lies on two different simple cycles, so it stays not desert.

Therefore, for each fixed `L` there exists a maximal `R(L)` such that the window `[L..R(L)]` is desert, and all `[L..R]` with `R ≤ R(L)` are also desert.

### Two pointers and counting formula
Maintain a sliding window `[L..R]` that is always desert. For each `L`:
1. Extend `R` as far as possible while keeping the window desert.
2. Once `R` is maximal, the number of valid pairs starting at `L` equals **`R − L + 1`**.
3. Add **`R − L + 1`** to the answer.
4. Remove edge `E_L` and increment `L`.

### Invariant: spanning forest with “one excluded edge per cycle”
Maintain a spanning forest of the current window graph.

In a cactus component, every simple cycle is represented by:
- exactly one **special** cycle edge that is **not** in the spanning tree, and
- all other edges of the cycle are tree edges marked `used = 1` (“this tree edge already belongs to a cycle”).

### Link-Cut Tree (LCT) edge-as-vertex modeling
Because an LCT stores data on vertices, represent each original edge `e_i = (u, v)` by creating a new vertex `mid(i)`.
Connect `u—mid(i)` and `mid(i)—v` in the LCT when the edge is in the spanning forest.

Store metadata on `mid(i)`:
- `idx(mid(i)) = i`
- `used(mid(i)) ∈ {0,1}`

Maintain path aggregates over edge-vertices:
- `minIdx` on a path (minimum `idx` among edge-vertices on the path)
- `maxUsed` on a path (maximum `used` among edge-vertices on the path)

Use lazy propagation to assign `used := 0` or `used := 1` on all edge-vertices along an LCT path.

### Insert edge `e_i = (u, v)` at the right boundary
Let `x = u`, `y = v`, and let `mid(i)` be the edge-vertex.

**Case A: `x` and `y` are disconnected.**
- Link `x—mid(i)—y` into the forest.
- No cycle is created.

**Case B: `x` and `y` are already connected.**
- The unique tree path `P(x, y)` plus the new edge forms one simple cycle.
- Query `maxUsed` on `P(x, y)`.
  - If `maxUsed = 1`, adding this edge would force an edge to lie on two cycles → window becomes invalid.
  - If `maxUsed = 0`, the cycle is allowed.

**Choosing the special excluded edge**
- The cycle edges are exactly: edges on `P(x, y)` plus the new edge `i`.
- Since edges are appended in increasing index order, `i` is the largest index on the cycle.
- Therefore the smallest-index cycle edge is the minimum `idx` on `P(x, y)`, obtainable by an LCT path-min query.

**Update**
- Cut that minimum-index edge-vertex from the spanning tree (it becomes the special excluded edge).
- Link `x—mid(i)—y` into the spanning tree.
- Apply a lazy path update `used := 1` on `P(x, y)` in the new tree.

### Delete edge `E_L` when moving the left boundary
Track for each edge whether it is currently linked in the spanning forest (tree edge) or excluded as the special non-tree edge.

- If `E_L` is the special excluded non-tree edge:
  - Deleting it destroys that cycle.
  - Apply a lazy path update `used := 0` on the tree path between its endpoints.

- If `E_L` is a tree edge:
  - Cut its edge-vertex `mid(L)` from the spanning forest.

### Complexity justification
Each insertion/deletion performs a constant number of LCT operations (path query, path update, link, cut).
Each LCT operation is `O(log N)` amortized.
Two pointers performs `O(M)` insertions and `O(M)` deletions total.
Total time: **`O(M log N)`**.


Prompt:
## Scenario
A corridor contains `n` unstable modules placed in a straight line, indexed from `1` to `n`.
Module `i` has an **activation radius** `a[i]`.

You may **manually trigger** a module only if it is still intact. Triggering module `i` instantly destroys every module `j` satisfying:

\[
|i - j| < a[i]
\]

Destroyed modules cannot be triggered later. Some modules are inert with `a[i] = 0`, which means they cannot be triggered at all.

Your goal is to destroy **all** modules using the **fewest manual triggers**, or determine that it cannot be done.

---

## Input
The first line contains an integer `t` — the number of test cases.

For each test case:
- The first line contains `n` — the number of modules.
- The second line contains `n` integers `a[1], a[2], ..., a[n]`.

### Constraints
- \(1 \le t \le 10^4\)
- \(2 \le n \le 5 \cdot 10^5\)
- \(0 \le a[i] \le n\)
- The sum of `n` across all test cases is at most \(5 \cdot 10^5\)

---

## Output
For each test case:

- Print `-1` if it is impossible to destroy all modules.

Otherwise print two lines:
1. An integer `k` — the minimum number of triggers.
2. `k` integers `p1 p2 ... pk` — a sequence of module indices to trigger, in the order you will trigger them.

If multiple optimal solutions exist, print any.

Does the response define the destroyed interval of triggering index i as Li = i - a[i] + 1 and Ri = i + a[i] - 1?

Does the response state that indices with a[i] = 0 cannot be triggered?

Does the response define dp[i] using Ri as the farthest-right destroyed position represented by state i?

Does the response specify Pattern A as a range-min query over endpoints x in [Li - 1, i - 1]?

Does the response specify Pattern B as a range-min query over endpoints x in [Li - 1, Ri] evaluated using only centers up to Li - 1?

Does the response include an offline scheduling mechanism that executes Pattern B queries at sweep time Li - 1?

Does the response perform a point update of the segment tree at position Ri with the value dp[i]?

Does the response select the final answer using min dp[i] over indices satisfying Ri >= n?

Does the response describe reconstruction using a predecessor pointer pre[i] obtained from the argmin returned by the segment tree query?

Does the response justify sorting the reconstructed triggers by nondecreasing a[i] using an adjacent-swap validity argument based on |x-y| >= a[x]?

## Key transformation
Triggering module `i` (only possible when `a[i] > 0`) destroys exactly the interval:

- \(L_i = i - a[i] + 1\)
- \(R_i = i + a[i] - 1\)

So each trigger corresponds to a closed range \([L_i, R_i]\).

The difficulty is that a module cannot be triggered after it has been destroyed, so the final set of triggered indices must admit some valid triggering order.

---

## DP state (compressed)
We only keep DP states where the **rightmost-triggered index is also the one that reaches farthest right**.

Let:

- `dp[i]` = minimum number of triggers needed so that, after executing them in some valid order, **all modules in `[1..R_i]` are destroyed**, and among the triggers used:
  - `i` is the **largest index**
  - `R_i` is the **maximum right endpoint** achieved

If `a[i] = 0`, then `dp[i] = +∞` since `i` cannot be triggered.

---

## Two valid transition patterns into `i`
Let \(L_i = i - a[i] + 1\) and note \(L_i - 1 = i - a[i]\).

We want to combine an earlier solution (represented by some endpoint `x`) with triggering `i` to achieve coverage up to `R_i`.

### Pattern A: trigger `i` after the earlier plan
To ensure `i` is still alive when it is triggered, the earlier plan must not destroy position `i`, which is guaranteed if the earlier plan’s farthest endpoint `x` satisfies:

- \(x < i\)

To ensure everything up to `L_i - 1` is already destroyed before triggering `i`, we need:

- \(x \ge L_i - 1\)

So we need:

- \(x \in [L_i - 1,\; i - 1]\)

### Pattern B: trigger `i` before finishing the earlier plan
If we trigger `i` earlier, we are allowed to use triggers strictly left of `L_i` later (their centers are not destroyed by `i` because `i` only destroys from `L_i` onwards).
So we can combine with an earlier plan that only uses centers in `[1..L_i-1]`, as long as that plan already guarantees destruction of `[1..L_i-1]`, meaning its endpoint `x` satisfies:

- \(x \ge L_i - 1\)

Also, we only need `x` up to `R_i` in this transition layer, because `dp[i]` stores states whose farthest-right reach is `R_i`:

- \(x \in [L_i - 1,\; R_i]\)

---

## Segment tree acceleration
Maintain a segment tree over endpoints `pos = 1..n`.

At any sweep moment, the segment tree stores at each `pos` the best pair:

- `(min_triggers_to_destroy_[1..pos], argmin_index)`

Whenever we have computed `dp[i]`, we perform a point update:

- `tree[R_i] = min(tree[R_i], (dp[i], i))`

This encodes: “there exists a plan with cost `dp[i]` whose farthest destroyed position is exactly `R_i` and whose representative rightmost trigger is `i`.”

### Getting transition values
For `dp[i]`, we need:

- `bestA = rangeMinQuery(L_i - 1, i - 1)` using all processed triggers up to `i-1`
- `bestB = rangeMinQuery(L_i - 1, R_i)` but **using only triggers with index ≤ (L_i - 1)**

To implement Pattern B, do it offline:
- For every `i`, schedule its Pattern B query to be executed at sweep time `time = L_i - 1`.
- Pattern A query can be executed at sweep time `time = i - 1`.

Let:
- `valA` be the numeric min from Pattern A query
- `valB` be the numeric min from Pattern B query

Then:
- `dp[i] = 1 + min(valA, valB)`
- If both are `+∞`, then `dp[i] = +∞`.

Base case:
- A plan of cost `0` that has destroyed nothing is represented by endpoint `0`.
  Practically, initialize the structure so that querying includes the possibility of starting fresh when `L_i <= 1`, by treating `endpoint 0` as cost `0` (often implemented via shifting indices or a special-case).

---

## Choosing the optimal finish
The whole line is destroyed if some chosen trigger `i` satisfies:

- \(R_i \ge n\)

Answer cost:
- \(\min_{i: R_i \ge n} dp[i]\)

If this minimum is `+∞`, print `-1`.

---

## Reconstruction
Store a predecessor pointer for each `i`:
- `pre[i]` = the `argmin_index` returned by the query that attained `min(valA, valB)` for `dp[i]`
- If the best came from the base case, use `pre[i] = 0`

Pick `best_end` = an index `i*` with minimal `dp[i]` among those with `R_i >= n`, then follow `pre[]` backwards to list all triggered indices.

---

## Output order (valid triggering sequence)
After reconstructing the set of indices, output them in **nondecreasing `a[i]`**.

Justification by swap argument:
- Consider two consecutive triggers `x` then `y` in a valid sequence with `a[x] > a[y]`.
- Since `y` was alive when triggered, `x` did not destroy `y`, so \(|x-y| \ge a[x]\).
- Then \(|x-y| \ge a[x] > a[y]\), so `y` would not destroy `x` either.
- Swapping `x` and `y` preserves validity.
- Repeating removes all inversions, producing a valid order sorted by `a[i]`.

---

## Complexity
- Segment tree operations: \(O(n \log n)\) per test file total across all tests due to sum of `n`
- Memory: \(O(n)\)


Prompt:
## Problem

There are `n` communication towers (numbered `1..n`) and `m` bidirectional wires.

Tower `i` accepts all integer frequencies in the interval `[l[i], r[i]]` (inclusive).

A tower `b` is called **accessible** from tower `a` if there exists an integer frequency `x` and a sequence of towers
`a = v1, v2, ..., vk = b`
such that:
- each consecutive pair `(vj, vj+1)` is connected by a wire, and
- **every** tower on the sequence accepts frequency `x` (i.e., `l[vj] ≤ x ≤ r[vj]` for all `j`).

Important: accessibility is **not transitive** because the frequency `x` may differ between different paths.

### Task
Output all towers that are accessible from tower `1`, in increasing order.

### Input
- First line: `n m` (`1 ≤ n ≤ 2⋅10^5`, `0 ≤ m ≤ 4⋅10^5`)
- Next `n` lines: `l[i] r[i]` (`1 ≤ l[i] ≤ r[i] ≤ 2⋅10^5`)
- Next `m` lines: undirected edges `v u` (`1 ≤ v,u ≤ n`, `v ≠ u`), no multi-edges.

### Output
Print the accessible tower indices from `1` in ascending order.

### Required sanity-checks (answer must also satisfy these)
Your approach must produce the following outputs on these additional testcases:

**Sanity Check A**
5 4
2 4
3 3
1 2
2 3
4 5
1 3
3 4
4 2
1 5

Expected output:
1 3 4 5

**Sanity Check B**
6 6
1 3
2 4
3 5
1 1
4 6
2 2
1 2
2 3
3 5
1 4
2 6
6 5

Expected output:
1 2 3 4 6

The answer explicitly restates that accessibility from tower 1 means: there exists a single integer frequency x such that all towers on the path accept x.

The answer reduces the problem to taking the union over frequencies x of the connected component of node 1 in the subgraph induced by towers active at x.

The answer defines the active interval of an edge (u,v) as [max(lu,lv), min(ru,rv)] and states the edge is usable exactly for x in that interval.

The answer uses an offline dynamic connectivity approach over x=1..MAX that adds each edge interval into a segment tree (or equivalent divide-and-conquer) over time.

The answer explicitly uses a DSU with rollback (persistent/undoable union-find) during the segment-tree traversal.

The answer states that at a leaf time x where tower 1 is inactive (x not in [l1,r1]), no towers should be marked/reported from that x.

The answer describes creating a merge-DAG (or DSU merge tree) where each union creates a new node with directed edges to the two merged component-nodes.

The answer states that a tower v is in the final output iff the singleton node for v is reachable from at least one marked component-node corresponding to find(1) at some x.

For the explicit input Sanity Check A: \"5 4\\n2 4\\n3 3\\n1 2\\n2 3\\n4 5\\n1 3\\n3 4\\n4 2\\n1 5\\n\" the answer states the output is exactly \"1 3 4 5\".

For the explicit input Sanity Check B: \"6 6\\n1 3\\n2 4\\n3 5\\n1 1\\n4 6\\n2 2\\n1 2\\n2 3\\n3 5\\n1 4\\n2 6\\n6 5\\n\" the answer states the output is exactly \"1 2 3 4 6\".

### Key reformulation: union over frequencies of the component of tower 1

Fix an integer frequency `x`. Let `Active(x) = { i : l[i] ≤ x ≤ r[i] }`, and consider the subgraph induced by `Active(x)`.
For this `x`, the towers accessible from `1` using frequency `x` are exactly the vertices in the connected component of `1`
in that induced subgraph (and if `1 ∉ Active(x)`, then nothing is accessible for that `x`).

So the final answer is the union over all `x` (from 1 to `MAX = 2⋅10^5`) of `Comp1(x)`.

This is an offline dynamic connectivity problem over time `x` where edges are active on intervals.

---

### Step 1: Convert vertex activity to edge activity intervals

An undirected wire `(u,v)` can be used at frequency `x` iff both endpoints accept `x`, i.e.:
- `x ∈ [l[u], r[u]] ∩ [l[v], r[v]]`

So the edge `(u,v)` is active on the (possibly empty) integer interval:
\[
I_{uv} = [ \max(l[u],l[v]),\ \min(r[u],r[v]) ].
\]
If `I_{uv}` is non-empty, then across all `x ∈ I_{uv}`, union-find should have `u` and `v` connected.

We also only care about times `x` where tower `1` is active, i.e. `x ∈ [l[1], r[1]]`, but it’s simplest to compute everything and ignore leaves where `1` is inactive.

---

### Step 2: Offline dynamic connectivity by divide-and-conquer + DSU rollback

Use a segment tree (or recursive divide-and-conquer) over the time axis `x ∈ [1..MAX]`.

For each edge `(u,v)` with active interval `[L,R] = I_{uv}`, add it to the segment tree nodes that fully cover `[L,R]`
(standard “add interval to segment tree” technique). Total inserts are `O(m log MAX)`.

Traverse the segment tree with a DSU that supports rollback:
- At each node, apply (union) all edges stored in that node.
- Recurse to children.
- Roll back DSU to the previous snapshot when returning.

At a leaf corresponding to a single frequency `x`, DSU represents connectivity of the active-edge graph at that `x`.

---

### Step 3: Record all components containing vertex 1 without enumerating vertices per leaf

Directly listing all vertices in the DSU component of `1` at each `x` is too slow.

Instead, build a **component-merge DAG** during DSU unions:

- Create a “component node” for every original vertex `v` (a leaf node representing the singleton component `{v}`).
- Whenever DSU unions two components with roots `ra` and `rb`, create a new DAG node `new`,
  add directed edges `new -> node(ra)` and `new -> node(rb)`,
  and set `node(newRoot) = new` as the representative DAG node for the merged DSU component.

This is a standard way to persist a history of merges; the DAG is a forest of merge trees (a “DSU merge tree”).

Now, whenever we are at a leaf time `x` such that tower `1` is active (`l[1] ≤ x ≤ r[1]`), we record the current DAG representative:
- `mark node( find(1) )`.

A tower `v` is accessible from `1` iff the singleton node for `v` is reachable from at least one marked node in this merge DAG.

So after processing all leaves, run a DFS/BFS from all marked nodes in the merge DAG and mark all reachable singleton nodes.
Output the corresponding vertex indices.

---

### Correctness sketch

- For any `x`, DSU at leaf `x` equals the connected components of the graph containing exactly edges whose `I_{uv}` covers `x`,
  which is exactly the induced subgraph on towers that accept `x` (edges exist iff both endpoints accept `x`).
- Thus the DSU component of `1` at leaf `x` is exactly `Comp1(x)`.
- Marking the DSU-component DAG node at each such `x` and then taking reachability down the merge DAG recovers the union of all
  vertices that ever belonged to `Comp1(x)` for some `x`.

---

### Complexity

- Building edge intervals: `O(n + m)`
- Segment tree interval insertion: `O(m log MAX)` with `MAX = 2⋅10^5`
- DSU rollback traversal: `O((m log MAX) α(n))` amortized (each inserted edge causes one union on each visited node)
- Merge DAG size: `O(n + number_of_unions)` = `O(n + m log MAX)`
- Final DFS on merge DAG: linear in DAG size.

This fits the constraints.

---

### Sanity checks (why the expected outputs match)

- Sanity A:
  - Frequency `x=2` connects 1–3–4, so {1,3,4} reachable.
  - Frequency `x=4` connects 1–5, so {1,5} reachable.
  - Union is {1,3,4,5}.

- Sanity B:
  - `x=1` gives {1,4}.
  - `x=2` gives {1,2,6}.
  - `x=3` gives {1,2,3}.
  - Union is {1,2,3,4,6}.


Machine Learning & Artificial Intelligence

Prompt: Can you prove that, for both (i) learning intervals on [0,1] under the uniform distribution and (ii) learning a linear separator in ℝᵈ under the uniform distribution on the unit sphere, any active learning algorithm requires Ω(d log(1/ε)) label requests to achieve error at most ε with probability ≥ 1/2 — and in particular, show that for each setting, there exists a distribution over target functions such that any algorithm making fewer than c·d log(1/ε) queries (for some constant c > 0) must output a hypothesis with error ≥ ε with probability at least 1/2?

Does the solution include a formula connecting query complexity k to packing size M (k ≥ log₂ M - 1)?

Does the solution explain that a deterministic algorithm with k queries can produce at most 2ᵏ distinct outputs?

Does the solution explain that each output hypothesis can cover at most one target in Θ?

Does the solution explicitly invoke Yao’s Minimax Principle?

Does the solution specify the grid spacing δ in terms of ε?

Does the solution show that the symmetric difference between any two intervals in Θ is ≥ δ > 2ε?

Does the solution apply the Packing Lemma to derive k = Ω(d log(1/ε)) for intervals?

Does the solution specify the angular separation γ ≈ ε needed to guarantee error > 2ε?

Does the solution conclude that any algorithm requires Ω(d log(1/ε)) queries for error ≤ ε with probability ≥ 1/2?

### Theorem
For both settings—(i) learning intervals on $[0,1]$ and (ii) learning linear separators in $\mathbb{R}^d$—there exists a distribution over target functions such that any active learning algorithm (randomized or deterministic) requires $\Omega(d \log(1/\varepsilon))$ label requests to achieve error $\le \varepsilon$ with probability $\ge 1/2$.

---

### 1. General Framework: The Packing Lower Bound

To prove a lower bound for *any* algorithm (including randomized ones), we use **Yao’s Minimax Principle**. It implies that the worst-case error probability of a randomized algorithm is lower-bounded by the expected error probability of the best *deterministic* algorithm against a specific "hard" input distribution $\mathcal{P}$.

We construct $\mathcal{P}$ as a uniform distribution over a finite set of "packed" hypotheses.

**Lemma (The Packing Bound):**
Let $\Theta = \{h_1, \dots, h_M\}$ be a finite set of hypotheses such that for any distinct pair $h_i, h_j \in \Theta$, the **disagreement probability under the data distribution** is $d(h_i, h_j) = P_{x \sim \mathcal{D}}(h_i(x) \neq h_j(x)) > 2\varepsilon$.
If the target $h^*$ is chosen uniformly from $\Theta$, then any deterministic active learning algorithm making fewer than $\lfloor \log_2 M \rfloor - 1$ queries will output a hypothesis with error $> \varepsilon$ with probability $\ge 1/2$.

**Proof of Lemma:**
1.  **Disjoint Error Regions:** Since the pairwise disagreement probability between targets is $> 2\varepsilon$, the error balls $B(h_i, \varepsilon) = \{h \mid d(h, h_i) \le \varepsilon\}$ are disjoint.
2.  **Injectivity:** Consequently, any single hypothesis $\hat{h}$ output by the learner can fall into the $\varepsilon$-ball of **at most one** target in $\Theta$.
3.  **Decision Tree Capacity:** A deterministic algorithm making $k$ binary queries can be modeled as a decision tree with at most $2^k$ leaves. Let $\mathcal{L}$ be the set of hypotheses output at these leaves.
4.  **Counting:** The algorithm succeeds if the true target $h^*$ corresponds to a leaf that "covers" it. Since $|\mathcal{L}| \le 2^k$ and each leaf covers at most 1 target, the algorithm can succeed on at most $2^k$ of the $M$ possible targets.
5.  **Probability:** Under the uniform distribution over $\Theta$, the probability of success is at most $2^k / M$. For the error probability to be $< 1/2$, we require $2^k / M > 1/2$, or $k > \log_2 M - 1$.

---

### 2. Part (i): Intervals on $[0,1]$

**Setting:**
*   Instance space $X = [0,1]$, Uniform distribution.
*   Hypothesis class $\mathcal{H} = \{ [a, b] \mid 0 \le a \le b \le 1 \}$.
*   Problem dimension $d=2$ (two parameters).

**Proof:**
We invoke the Packing Lemma by constructing a set $\Theta$ of size $\approx (1/\varepsilon)^2$.

1.  **Construction:**
    Let $\delta = 3\varepsilon$. We define a grid of possible endpoints:
    *   Left endpoints $L = \{ 0, \delta, 2\delta, \dots \} \cap [0, 1/2]$.
    *   Right endpoints $R = \{ 1, 1-\delta, 1-2\delta, \dots \} \cap [1/2, 1]$.
    *   Let $\Theta = \{ [l, r] \mid l \in L, r \in R \}$.
    *   *Note:* The restriction to $[0, 1/2]$ and $[1/2, 1]$ ensures $l \le r$ for all pairs, making all intervals valid.

2.  **Separation:**
    For any two distinct intervals $h_1 = [l_1, r_1]$ and $h_2 = [l_2, r_2]$, the disagreement probability (measure of the symmetric difference) is:
    $$ d(h_1, h_2) = |l_1 - l_2| + |r_1 - r_2| $$
    Since $h_1 \neq h_2$, they must differ by at least one grid step $\delta$ in either the left or right coordinate. Thus, $d(h_1, h_2) \ge \delta = 3\varepsilon > 2\varepsilon$.

3.  **Cardinality ($M$):**
    The size of $L$ is $\approx \frac{1/2}{3\varepsilon} \propto \frac{1}{\varepsilon}$. Similarly, $|R| \propto \frac{1}{\varepsilon}$.
    $$ M = |L| \times |R| = \Omega\left(\frac{1}{\varepsilon^2}\right) $$

4.  **Lower Bound:**
    Applying the Lemma:
    $$ k \ge \log_2\left( \frac{c}{\varepsilon^2} \right) \approx 2 \log_2\left(\frac{1}{\varepsilon}\right) $$
    Since $d=2$, this confirms **$k = \Omega(d \log(1/\varepsilon))$**.

---

### 3. Part (ii): Linear Separators in $\mathbb{R}^d$

**Setting:**
*   Instance space $X = \mathbb{S}^{d-1}$ (unit sphere), Uniform surface measure.
*   Hypothesis class $\mathcal{H} = \{ x \mapsto \text{sign}(w \cdot x) \}$.
*   Problem dimension is $d$.

**Proof:**
We invoke the Packing Lemma by constructing a spherical code $\Theta$ of size roughly $(1/\varepsilon)^{d-1}$.

1.  **Geometry of Error:**
    For unit vectors $u, v$, the disagreement probability is exactly the normalized surface measure of the wedge between the hyperplanes defined by $u$ and $v$. This is proportional to the angle:
    $$ d(h_u, h_v) = \frac{\theta(u, v)}{\pi} $$
    To ensure separation $> 2\varepsilon$, we require angular separation $\theta > 2\pi\varepsilon$.

2.  **Construction (Spherical Packing):**
    We need a set $\Theta \subset \mathbb{S}^{d-1}$ such that every pair is separated by angle $\gamma \approx \varepsilon$. The maximum size of such a set is bounded by comparing the total surface area of the sphere to the area of a spherical cap of radius $\gamma/2$.
    Using standard bounds for spherical codes (e.g., Gilbert-Varshamov or Kabatiansky-Levenshtein), the packing number scales as:
    $$ M = \Omega\left( \left(\frac{1}{\gamma}\right)^{d-1} \right) = \Omega\left( \left(\frac{1}{\varepsilon}\right)^{d-1} \right) $$
    *(This holds up to constant factors in the exponent).*

3.  **Lower Bound:**
    Applying the Lemma:
    $$ k \ge \log_2 M - 1 \approx (d-1) \log_2 \left(\frac{1}{\varepsilon}\right) $$
    For $d \ge 2$, we have $d-1 \ge d/2$. Thus:
    $$ k = \Omega\left( d \log \frac{1}{\varepsilon} \right) $$

### Conclusion

In both cases, we established that the set of distinguishable hypotheses has a size $M$ that scales as $(1/\varepsilon)^{\Theta(d)}$. Since active learning queries provide at most 1 bit of information per query, identifying the correct target requires $\log_2 M$ queries.
Therefore, any algorithm requires $\Omega(d \log(1/\varepsilon))$ queries.


Prompt:
You are given two positive integers **n** and **k**. For every integer **x** such that
1 <= x <= n, write the representation of **x** in base **k** (without leading zeros).

Each representation is a sequence of digits. Collect all n such digit-sequences and sort
them in **lexicographic order**: two sequences are compared digit by digit from left to
right (each digit compared by its integer value); at the first position where they differ,
the sequence with the smaller digit comes first; if one sequence is a proper prefix of
the other, the shorter one comes first.

After sorting, index the array from 1 to n (1-based indexing).

**Your task:** For each test case, compute how many integers i (1 <= i <= n) satisfy the
following condition: the base-k representation of i appears **exactly at position i** in
the sorted array.

**Additionally, describe the algorithm you use and explain why it runs efficiently for
the given constraints.**

---

### Input

5
7 2
10 3
100 10
1000000000000000000 10
1000000000000000000 2

---

The number 5 inside the `Input` represents the number of test cases.

### Notes

- The representation of 1 in any base is the single digit [1].
- The number 7 in base 3 is [2, 1].
- The number 81 in base 9 is [1, 0, 0].
- For the first test case (n=7, k=2), the base-2 representations sorted
  lexicographically are:
  [1], [1,0], [1,0,0], [1,0,1], [1,1], [1,1,0], [1,1,1]
  corresponding to decimal values 1, 2, 4, 5, 3, 6, 7.
  The values i where position equals i are: i=1 (value 1 at position 1),
  i=2 (value 2 at position 2), i=6 (value 6 at position 6),
  i=7 (value 7 at position 7). So the answer is 4.

---

### Output

For each test case, output a single integer on its own line: the count of values i such
that the base-k representation of i is located at position i in the lexicographically
sorted array.

The response outputs 4 as the answer for the first test case (n=7, k=2).

The response outputs 1 as the answer for the second test case (n=10, k=3).

The response outputs 11 as the answer for the third test case (n=100, k=10).

The response outputs 11 as the answer for the fourth test case (n=10^18, k=10).

The response outputs 2 as the answer for the fifth test case (n=10^18, k=2).

The response describes a method for computing the lexicographic rank of a number by counting how many valid numbers (at most n) are lexicographically smaller, using prefix-based or subtree-based counting.

The response explains why the fixed-point solutions within each group of same-length representations form a contiguous range (or equivalently, why binary search is applicable), by arguing that the difference between lexicographic rank and numeric rank is non-decreasing within such a group.

The response correctly states or demonstrates that the algorithm runs in polynomial-in-log-n time (e.g., O((log n)^2) or O((log n)^3)) per test case, making it feasible for n up to 10^18.

PROBLEM ANALYSIS
================

We must count how many x in [1..n] satisfy: lexicographic_rank(x) = x,
where lexicographic_rank(x) is the 1-based position of x's base-k
representation in the sorted array of all representations from 1 to n.

KEY INSIGHT: TRIE MODEL
========================

Model all base-k representations as paths in a digit trie:
- Root has children for digits 1..k-1 (no leading zeros)
- Each internal node has children for digits 0..k-1
- A node is "active" if the number it represents is <= n

Two traversal orders correspond to two rankings:
- BFS (level-order): visits by increasing length, then by numeric value
  within equal length. This gives bfs_rank(x) = x for all x.
- DFS (pre-order, children in digit order 0..k-1): visits in
  lexicographic order. This gives dfs_rank(x) = lexicographic_rank(x).

Therefore we need: count of x where dfs_rank(x) = x.

COMPUTING dfs_rank(x)
=====================

Given x with base-k digits d_1 d_2 ... d_L:

dfs_rank(x) = 1 + (number of valid entries < x in lex order)

Algorithm:
  rank = 0, prefix = 0
  For each digit position i (0-indexed):
    For every digit c from start_digit to d_i - 1:
      rank += subtree_size(prefix * k + c, k, n)
    prefix = prefix * k + d_i
    If this is not the last digit and prefix <= n:
      rank += 1   (count the prefix itself as a valid number)
  Return rank + 1

where start_digit = 1 for the first position, 0 otherwise.

SUBTREE SIZE (prefix counting)
==============================

Count of numbers in [1..n] whose base-k representation starts with
a given prefix:

  def subtree_size(prefix, k, n):
      low, high, count = prefix, prefix + 1, 0
      while low <= n:
          count += min(n + 1, high) - low
          low *= k
          high *= k
      return count

For summing over consecutive digits c in [a, b), use:
  subtree_range_size(prefix*k + a, prefix*k + b, k, n)
which avoids iterating over each digit individually (critical for large k).

MONOTONICITY AND BINARY SEARCH
==============================

Define f(x) = dfs_rank(x) - x.

Within each "layer" (set of numbers with the same base-k representation
length L), f(x) is non-decreasing. Proof: for consecutive x, x+1 in the
same layer, bfs_rank increases by 1 while dfs_rank increases by >= 1.

Consequence: the set of solutions {x : f(x) = 0} within each layer
forms a contiguous (possibly empty) range. We can binary search for the
first and last x satisfying f(x) = 0 in each layer.

FULL ALGORITHM
==============

For each test case (n, k):
  answer = 0
  For each digit-length L = 1, 2, ...:
    Compute layer range [lo, hi] where lo = k^(L-1), hi = min(k^L - 1, n)
    (For L=1: lo=1, hi=min(k-1, n))
    If lo > n: break
    Binary search in [lo, hi] for first x with dfs_rank(x) >= x
    Binary search in [lo, hi] for last x with dfs_rank(x) <= x
    If valid range found: answer += last - first + 1
  Output answer

COMPLEXITY: O((log_k n)^2 * log(layer_size)) per test case.
  - Number of layers: O(log_k n)
  - Binary search per layer: O(log(layer_size))
  - dfs_rank: O(log_k n) with range subtree counting
This handles n, k up to 10^18 efficiently.

VERIFICATION
============

Working Python solution:

  def to_digits(x, k):
      digits = []
      while x > 0:
          digits.append(x % k)
          x //= k
      return digits[::-1]

  def count_range_subtrees(lo_pref, hi_pref, k, n):
      low, high, count = lo_pref, hi_pref, 0
      while low <= n:
          count += min(n + 1, high) - low
          low *= k
          high *= k
      return count

  def lex_rank(x, k, n):
      digits = to_digits(x, k)
      rank, prefix = 0, 0
      for i, d in enumerate(digits):
          start = 1 if i == 0 else 0
          if d > start:
              rank += count_range_subtrees(prefix*k + start, prefix*k + d, k, n)
          prefix = prefix * k + d
          if i < len(digits) - 1 and prefix <= n:
              rank += 1
      return rank + 1

  def solve(n, k):
      count, L = 0, 1
      while True:
          if L == 1:
              lo, hi = 1, min(k - 1, n)
          else:
              lo = k ** (L - 1)
              if lo > n: break
              hi = min(lo * k - 1, n)
          if lo > hi:
              L += 1; continue
          a, b = lo, hi; first = hi + 1
          while a <= b:
              mid = (a + b) // 2
              if lex_rank(mid, k, n) >= mid: first, b = mid, mid - 1
              else: a = mid + 1
          a, b = lo, hi; last = lo - 1
          while a <= b:
              mid = (a + b) // 2
              if lex_rank(mid, k, n) <= mid: last, a = mid, mid + 1
              else: b = mid - 1
          if first <= last: count += last - first + 1
          L += 1
      return count

Verified against brute force for hundreds of random (n, k) pairs with
n up to 500 and k up to 30.

ANSWERS
=======

Test case 1: n=7, k=2           =>  4
Test case 2: n=10, k=3          =>  1
Test case 3: n=100, k=10        =>  11
Test case 4: n=10^18, k=10      =>  11
Test case 5: n=10^18, k=2       =>  2

Output:
4
1
11
11
2
```