#include <bits/stdc++.h>
using namespace std;

struct TestCase {
    int n;
    long long k;
    vector<vector<long long>> A;
    long long answer;
};

mt19937_64 rng_gen(42);

TestCase gen_matrix(int n, long long k, function<long long(int,int)> valfn) {
    TestCase tc;
    tc.n = n; tc.k = k;
    tc.A.assign(n+1, vector<long long>(n+1, 0));
    vector<long long> all;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++) {
            tc.A[i][j] = valfn(i, j);
            all.push_back(tc.A[i][j]);
        }
    sort(all.begin(), all.end());
    tc.answer = all[k-1];
    return tc;
}

TestCase gen_multiplicative(int n, long long k) { return gen_matrix(n, k, [](int i, int j) -> long long { return (long long)i * j; }); }
TestCase gen_shifted(int n, long long k) { return gen_matrix(n, k, [n](int i, int j) -> long long { return (long long)(i + n) * (j + n); }); }
TestCase gen_additive(int n, long long k) { return gen_matrix(n, k, [](int i, int j) -> long long { return i + j; }); }
TestCase gen_random_sorted(int n, long long k) {
    TestCase tc;
    tc.n = n; tc.k = k;
    tc.A.assign(n+1, vector<long long>(n+1, 0));
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            tc.A[i][j] = (long long)i * 1000000 + (long long)j * 1000 + (rng_gen() % 500);
    for (int i = 1; i <= n; i++)
        for (int j = 2; j <= n; j++)
            tc.A[i][j] = max(tc.A[i][j], tc.A[i][j-1]);
    for (int j = 1; j <= n; j++)
        for (int i = 2; i <= n; i++)
            tc.A[i][j] = max(tc.A[i][j], tc.A[i-1][j]);
    vector<long long> all;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            all.push_back(tc.A[i][j]);
    sort(all.begin(), all.end());
    tc.answer = all[k-1];
    return tc;
}

struct Solver {
    const TestCase& tc;
    int query_count;
    vector<long long> memo;
    int n;

    Solver(const TestCase& t) : tc(t), query_count(0), n(t.n) {
        memo.assign(2002 * 2002, -1);
    }

    long long do_query(int r, int c) {
        int key = r * 2001 + c;
        if (memo[key] != -1) return memo[key];
        query_count++;
        memo[key] = tc.A[r][c];
        return memo[key];
    }

    long long solve() {
        long long k = tc.k;
        long long N2 = (long long)n * n;

        if (n == 1) return do_query(1, 1);

        long long heap_k = min(k, N2 - k + 1);
        if (heap_k + n <= 24000) {
            if (k <= N2 - k + 1) {
                priority_queue<tuple<long long, int, int>, vector<tuple<long long, int, int>>, greater<>> pq;
                vector<vector<bool>> vis(n + 1, vector<bool>(n + 1, false));
                pq.emplace(do_query(1, 1), 1, 1);
                vis[1][1] = true;
                long long result = -1;
                for (long long i = 0; i < k; i++) {
                    auto [v, r, c] = pq.top(); pq.pop();
                    result = v;
                    if (r + 1 <= n && !vis[r + 1][c]) { vis[r + 1][c] = true; pq.emplace(do_query(r + 1, c), r + 1, c); }
                    if (c + 1 <= n && !vis[r][c + 1]) { vis[r][c + 1] = true; pq.emplace(do_query(r, c + 1), r, c + 1); }
                }
                return result;
            } else {
                long long kk = N2 - k + 1;
                priority_queue<tuple<long long, int, int>> pq;
                vector<vector<bool>> vis(n + 1, vector<bool>(n + 1, false));
                pq.emplace(do_query(n, n), n, n);
                vis[n][n] = true;
                long long result = -1;
                for (long long i = 0; i < kk; i++) {
                    auto [v, r, c] = pq.top(); pq.pop();
                    result = v;
                    if (r - 1 >= 1 && !vis[r - 1][c]) { vis[r - 1][c] = true; pq.emplace(do_query(r - 1, c), r - 1, c); }
                    if (c - 1 >= 1 && !vis[r][c - 1]) { vis[r][c - 1] = true; pq.emplace(do_query(r, c - 1), r, c - 1); }
                }
                return result;
            }
        }

        vector<int> L(n + 1, 1), R(n + 1, n);
        long long k_rem = k;

        // Store previous boundary values for adaptive pivot selection
        vector<long long> prev_boundary_vals;
        vector<int> prev_boundary_weights;

        for (int iter = 0; iter < 100; iter++) {
            vector<int> active;
            long long total_cand = 0;
            for (int i = 1; i <= n; i++) {
                if (L[i] <= R[i]) {
                    active.push_back(i);
                    total_cand += R[i] - L[i] + 1;
                }
            }
            int na = active.size();
            if (total_cand == 0) break;
            if (total_cand == 1) {
                for (int i : active) return do_query(i, L[i]);
                break;
            }

            long long budget = 49500 - query_count;
            if (k_rem + na <= budget) {
                priority_queue<tuple<long long, int, int>, vector<tuple<long long, int, int>>, greater<>> pq;
                for (int i : active) pq.emplace(do_query(i, L[i]), i, L[i]);
                for (long long t = 1; t < k_rem; t++) {
                    auto [v, r, c] = pq.top(); pq.pop();
                    if (c + 1 <= R[r]) pq.emplace(do_query(r, c + 1), r, c + 1);
                }
                return get<0>(pq.top());
            }
            long long rev_k = total_cand - k_rem + 1;
            if (rev_k + na <= budget) {
                priority_queue<tuple<long long, int, int>> pq;
                for (int i : active) pq.emplace(do_query(i, R[i]), i, R[i]);
                for (long long t = 1; t < rev_k; t++) {
                    auto [v, r, c] = pq.top(); pq.pop();
                    if (c - 1 >= L[r]) pq.emplace(do_query(r, c - 1), r, c - 1);
                }
                return get<0>(pq.top());
            }

            // Pivot selection: sample from each active row at target_frac position
            // But also use midpoint of each row (2 samples per row) for better coverage
            double target_frac = (double)(k_rem - 0.5) / total_cand;

            // Strategy: sample from a subset of rows, weighted by width
            // For rows with width > median_width, sample at both target_frac and 0.5 positions
            vector<pair<long long, int>> vw; // (value, weight=width)

            // Sample every active row at the target_frac position
            int sample_step = max(1, na / min(na, max(1, (int)ceil(sqrt((double)na) * 4))));
            for (int idx = 0; idx < na; idx += sample_step) {
                int i = active[idx];
                int width = R[i] - L[i] + 1;
                int col = L[i] + (int)(target_frac * (width - 1));
                col = max(L[i], min(R[i], col));
                vw.push_back({do_query(i, col), width});
            }

            // Sort by value and find weighted quantile
            sort(vw.begin(), vw.end());
            long long total_w = 0;
            for (auto& [v, w] : vw) total_w += w;
            long long target_w = (long long)(target_frac * total_w);
            long long cum = 0;
            long long pivot = vw[vw.size() / 2].first;
            for (auto& [v, w] : vw) {
                cum += w;
                if (cum >= target_w) {
                    pivot = v;
                    break;
                }
            }

            // Staircase walk (bottom-to-top)
            vector<int> p_le(n + 1, 0);
            {
                int j = 0;
                for (int idx = na - 1; idx >= 0; idx--) {
                    int i = active[idx];
                    j = max(j, L[i]);
                    while (j <= R[i] && do_query(i, j) <= pivot) j++;
                    p_le[i] = j - 1;
                }
            }

            long long cle = 0;
            for (int i : active) {
                int rl = min(p_le[i], R[i]);
                if (rl >= L[i]) cle += rl - L[i] + 1;
            }

            // Store boundary values for potential use in next iteration
            prev_boundary_vals.clear();
            prev_boundary_weights.clear();

            if (cle >= k_rem) {
                for (int i : active) {
                    if (p_le[i] >= L[i]) {
                        // Boundary: a[i][p_le[i]] is the max value in the kept region for row i
                        prev_boundary_vals.push_back(do_query(i, p_le[i])); // cached
                        prev_boundary_weights.push_back(min(p_le[i], R[i]) - L[i] + 1);
                    }
                    R[i] = min(R[i], p_le[i]);
                }
            } else {
                k_rem -= cle;
                for (int i : active) {
                    if (p_le[i] + 1 <= R[i]) {
                        prev_boundary_vals.push_back(do_query(i, p_le[i] + 1)); // boundary of > side, cached from walk
                        prev_boundary_weights.push_back(R[i] - max(L[i], p_le[i] + 1) + 1);
                    }
                    L[i] = max(L[i], p_le[i] + 1);
                }
            }
        }
        return -1;
    }
};

int main() {
    struct TestDef { string name; function<TestCase()> gen; };
    vector<TestDef> tests;
    tests.push_back({"additive n=100 k=5000", []{ return gen_additive(100, 5000); }});
    tests.push_back({"mult n=100 k=5000", []{ return gen_multiplicative(100, 5000); }});
    tests.push_back({"additive n=500 k=125000", []{ return gen_additive(500, 125000); }});
    tests.push_back({"mult n=500 k=125000", []{ return gen_multiplicative(500, 125000); }});
    tests.push_back({"random n=500 k=125000", []{ return gen_random_sorted(500, 125000); }});
    tests.push_back({"additive n=2000 k=2000000", []{ return gen_additive(2000, 2000000); }});
    tests.push_back({"mult n=2000 k=2000000", []{ return gen_multiplicative(2000, 2000000); }});
    tests.push_back({"shifted n=2000 k=2000000", []{ return gen_shifted(2000, 2000000); }});
    tests.push_back({"random n=2000 k=2000000", []{ return gen_random_sorted(2000, 2000000); }});
    tests.push_back({"mult n=2000 k=100000", []{ return gen_multiplicative(2000, 100000); }});
    tests.push_back({"mult n=2000 k=3900000", []{ return gen_multiplicative(2000, 3900000); }});

    for (auto& t : tests) {
        auto tc = t.gen();
        Solver s(tc);
        long long result = s.solve();
        bool correct = (result == tc.answer);
        int used = s.query_count;
        double score = !correct ? 0.0 : (used <= tc.n ? 1.0 : (used >= 50000 ? 0.0 : (50000.0 - used) / (50000.0 - tc.n)));
        printf("%-45s q=%6d %s score=%.4f\n", t.name.c_str(), used, correct ? "OK" : "WRONG", score);
    }
}