<|instruction|>
Solve the below problem. The solution should start with an abridged problem statement. Then key observations. Then full solution based on the observations. Then C++ and Python implementations with comments.

448. Controlled Tournament
Time limit per test: 1 second(s)
Memory limit: 262144 kilobytes
input: standard
output: standard

National Association of Tennis is planning to hold a tennis competition among professional players. The competition is going to be a knockout tournament, and you are assigned the task to make the arrangement of players in the tournament. You are given the detailed report about all participants of the competition. The report contains the results of recent matches between all pairs of the participants. Examining the data, you've noticed that it is only up to the opponent whether one player wins or not. Since one of your special friends are attending the competition, you want him to get the best prize. So you want to know the possibility where he wins the gold medal. However it is not so easy to figure out because there are many participants. You have decided to write a program which calculates the number of possible arrangements of tournament in which your friend wins the gold medal. In order to make your trick hidden from everyone, you need to avoid making a factitive tournament tree. So you have to consider only such tournaments that the height of your tournament tree is minimal possible.
Input
The input has the format as described below.
N M
R11 R12... R1N
R21 R22... R2N
...
RN1 RN2... RNN
N is the number of players (1 ≤ N ≤ 16), and M is your friend's ID (numbered from 1). Rij is the result of a match between the i-th player and the j-th player. When i-th player always wins, Rij = 1. Otherwise, Rij = 0. It is guaranteed that the matrix is consistent: for all i != j, Rij = 0 if and only if Rji = 1. The diagonal elements Rii are just given for convenience and are always 0.
Output
Your program should output in a line the number of possible tournaments in which your friend wins the first prize.
Example(s)
sample input
sample output
2 1
0 1
0 0
1

sample input
sample output
2 1
0 0
1 0
0

sample input
sample output
3 3
0 1 1
0 0 1
0 0 0
0

sample input
sample output
3 3
0 1 0
0 0 0
1 1 0
3

sample input
sample output
3 1
0 1 0
0 0 0
1 1 0
0

sample input
sample output
3 3
0 1 0
0 0 1
1 0 0
1

sample input
sample output
4 1
0 0 0 1
1 0 0 1
1 1 0 0
0 0 1 0
0

sample input
sample output
6 4
0 0 0 0 0 1
1 0 1 0 1 0
1 0 0 1 1 0
1 1 0 0 1 0
1 0 0 0 0 0
0 1 1 1 1 0
11

sample input
sample output
7 2
0 1 0 0 0 1 0
0 0 1 0 1 1 1
1 0 0 1 1 0 0
1 1 0 0 0 1 0
1 0 0 1 0 0 1
0 0 1 0 1 0 0
1 0 1 1 0 1 0
139

sample input
sample output
8 6
0 0 0 0 1 0 0 0
1 0 1 1 0 0 0 0
1 0 0 0 1 0 0 0
1 0 1 0 0 1 0 1
0 1 0 1 0 0 1 0
1 1 1 0 1 0 0 1
1 1 1 1 0 1 0 0
1 1 1 0 1 0 1 0
78

<|response|>
1. Abridged problem statement

Given N players (N≤16) and a win matrix R where R[i][j]=1 means player i always beats j, count the number of ways to fill a knockout tournament bracket of minimal height T=⌈log₂N⌉ (with byes if needed) so that a designated player M wins the championship. Each leaf of the perfect binary tree of height T is assigned either one player or left empty (a bye). Matches proceed up the tree, and whenever two players meet, the matrix R determines the winner.

2. Key observations

- Minimal height T=⌈log₂N⌉ implies exactly 2ᵀ leaves. Some leaves are empty (byes), but the total number of actual players in leaves is N.
- A subtree that contains k real players must have height at least ⌈log₂k⌉. We only count arrangements that use exactly the minimal height possible to place all N players.
- We need to count ways to choose which k players go into the left subtree, which into the right, and ensure that the known match outcomes lead to a specific winner.
- A direct enumeration over subsets and masks is O(3ᴺ) or worse. We can speed up the "sum over all partitions of mask into mask₁∪mask₂" by using the Fast Walsh–Hadamard transform (FWT) for subset convolution (XOR-convolution trick).

3. Full solution approach

a. Definitions and DP state
   - Let FULL = (1≪N)–1 be the bitmask of all players.
   - Precompute best_size[k] = ⌈log₂k⌉, the minimal rounds needed for k players (best_size[k] = best_size[k>>1] + 1).
   - Let steps = best_size[N-1] + 1 be the total number of rounds.
   - We define dp[step][winner][size] as a vector of length 2ᴺ in FWT-domain, where:
       * step = number of rounds used so far (0≤step<steps)
       * winner = champion of this subtree (0≤winner<N)
       * size = the number of players in the subtree (1≤size≤N)
     and dp[step][winner][size][mask] (in ordinary domain) = number of ways to choose exactly the set of players = mask (popcount(mask)=size) in a subtree of the given round count that produces the given winner, using minimal height constraints for all sub-subtrees.
   - We keep all dp vectors in FWT-domain so that merging two subtrees is an elementwise multiply.

b. Base case
   - For each player i (0≤i<N), dp[0][i][1] has exactly one way to place i alone in a single-player subtree (mask = 1≪i). In FWT-domain this is a length-2ᴺ vector with a single 1 at index (1≪i), then transformed by xor_transform.

c. Transitions (merging two subtrees)
   - Consider splitting a subtree of size k into sizes sz_a and sz_b with sz_a+sz_b=k.
   - Let step_a and step_b be rounds used by the left and right subtrees respectively; to form the parent we need new_step = max(step_a, step_b)+1. We only allow step_a≥best_size[sz_a-1] and step_b≥best_size[sz_b-1].
   - For every possible left-winner x and right-winner y (with y > x to avoid double-counting), we know the final winner w = x if R[x][y]=1 else y. We also handle the mirrored (y, x) case implicitly via the loop y > x and then handling R[y][x] if needed — actually the current code only iterates y > x and uses R[x][y] to determine the winner; by symmetry each ordered pair is covered.
   - In FWT-domain we simply do:
       dp[new_step][winner][sz_a+sz_b] += dp[step_a][x][sz_a] * dp[step_b][y][sz_b]  (elementwise)
     accumulated via multiply_add, which does 4-element chunked SIMD-friendly unrolling.
   - A pruning check skips cases where remaining players can't be scheduled in remaining steps.

d. Final answer
   - After filling dp, look at dp[steps-1][M][N], inverse-FWT it, and read the coefficient of FULL. That is the count of ways to arrange all N players so that M wins.

e. Complexity
   - Number of steps ≤ 5 (since N≤16, steps = best_size[15]+1 = 5).
   - Each dp entry is a length 2ᴺ (≤65536) vector.
   - Each FWT or inverse-FWT costs O(N·2ᴺ). Each merge is O(2ᴺ).
   - Total is a few hundred passes over vectors of length ≤65536—feasible in optimized C++ with SIMD and unroll hints (pragmas).

4. C++ Solution

```cpp
#include <bits/stdc++.h>

#pragma GCC optimize("O3")
#pragma GCC target("avx2")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("tree-vectorize")

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

int n, m;
vector<vector<int>> R;

void multiply_add(
    vector<uint64_t>& result, const vector<uint64_t>& a,
    const vector<uint64_t>& b
) {
    const size_t size = a.size();
    const size_t chunk_size = 4;
    const size_t chunk_end = size - (size % chunk_size);

    for(size_t i = 0; i < chunk_end; i += chunk_size) {
        result[i] += a[i] * b[i];
        result[i + 1] += a[i + 1] * b[i + 1];
        result[i + 2] += a[i + 2] * b[i + 2];
        result[i + 3] += a[i + 3] * b[i + 3];
    }

    for(size_t i = chunk_end; i < size; i++) {
        result[i] += a[i] * b[i];
    }
}

void read() {
    cin >> n >> m;
    m--;
    R.resize(n, vector<int>(n));
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < n; j++) {
            cin >> R[i][j];
        }
    }
}

void xor_transform(vector<uint64_t>& a, bool reverse = false) {
    int n = a.size();
    for(int i = 1; i < n; i <<= 1) {
        for(int j = 0; j < n; j += (i << 1)) {
            for(int k = 0; k < i; k++) {
                uint64_t x = a[j + k];
                uint64_t y = a[j + k + i];
                a[j + k] = x + y;
                a[j + k + i] = x - y;
            }
        }
    }

    if(reverse) {
        for(int i = 0; i < n; i++) {
            a[i] /= n;
        }
    }
}

void solve() {
    // We count minimal-height knockout tournaments in which player m wins. A
    // tournament is built by merging two disjoint sub-brackets: the winners of
    // two sub-tournaments over disjoint sets of players play, and the match
    // result R decides who advances. The bracket height must be minimal, which
    // means a sub-bracket over s players needs exactly ceil(log2(s)) rounds;
    // best_size[s-1] is that minimum number of merge rounds.
    //
    // The DP state is dp[step][winner][size][mask]: the number of ways to form
    // a sub-bracket of the given size and round count whose champion is
    // `winner` and whose participant set is `mask`. Merging two sub-brackets
    // requires their masks to be disjoint and their union to be tracked, which
    // is an XOR (subset) convolution over the participant bitmask.
    //
    // To make the convolution cheap we keep every dp[...] array in the
    // Walsh-Hadamard transformed domain over the n-bit mask space. There,
    // combining two sub-brackets is just pointwise multiplication
    // (multiply_add accumulates a[i] * b[i] for the winner determined by R).
    // Leaves dp[0][i][1] start as a single-player bracket for player i (the
    // indicator of mask {i}), transformed once. We iterate sub-bracket size
    // pairs in increasing max-size order so all inputs are ready, only allowing
    // round counts that keep the final tree at minimal height. Finally we
    // inverse-transform dp[steps-1][m][n] and read off the coefficient for the
    // full participant set (1 << n) - 1.

    vector<int> best_size(n + 1, 0);
    for(int i = 1; i <= n; i++) {
        best_size[i] = best_size[i >> 1] + 1;
    }

    int steps = best_size[n - 1] + 1;
    vector<vector<vector<vector<uint64_t>>>> dp(
        steps, vector<vector<vector<uint64_t>>>(
                   n, vector<vector<uint64_t>>(n + 1, vector<uint64_t>())
               )
    );

    for(int i = 0; i < n; i++) {
        dp[0][i][1].assign(1 << n, 0);
        dp[0][i][1][1 << i] = 1;
        xor_transform(dp[0][i][1]);
    }

    vector<pair<int, int>> sz_a_b;
    for(int sz_a = 1; sz_a <= n; sz_a++) {
        for(int sz_b = 1; sz_b <= n; sz_b++) {
            if(sz_a + sz_b > n) {
                continue;
            }
            sz_a_b.emplace_back(sz_a, sz_b);
        }
    }

    sort(sz_a_b.begin(), sz_a_b.end(), [](const auto& a, const auto& b) {
        return max(a.first, a.second) < max(b.first, b.second);
    });

    for(auto [sz_a, sz_b]: sz_a_b) {
        for(int step_a = 0; step_a + 1 < steps; step_a++) {
            if(step_a < best_size[sz_a - 1]) {
                continue;
            }
            for(int step_b = 0; step_b + 1 < steps; step_b++) {
                if(step_b < best_size[sz_b - 1]) {
                    continue;
                }
                for(int x = 0; x < n; x++) {
                    for(int y = x + 1; y < n; y++) {
                        int winner = R[x][y] ? x : y;
                        int new_step = max(step_a, step_b) + 1;
                        auto& dp_a = dp[step_a][x][sz_a];
                        auto& dp_b = dp[step_b][y][sz_b];
                        auto& dp_winner = dp[new_step][winner][sz_a + sz_b];

                        if(dp_a.empty() || dp_b.empty()) {
                            continue;
                        }

                        int left = n - sz_a - sz_b;
                        int left_steps = steps - new_step - 1;
                        if(left != 0 && left_steps == 0) {
                            continue;
                        }

                        if(dp_winner.empty()) {
                            dp_winner.assign(1 << n, 0);
                        }

                        multiply_add(
                            dp[new_step][winner][sz_a + sz_b],
                            dp[step_a][x][sz_a], dp[step_b][y][sz_b]
                        );
                    }
                }
            }
        }
    }

    if(dp[steps - 1][m][n].empty()) {
        cout << 0 << '\n';
        return;
    }
    xor_transform(dp[steps - 1][m][n], true);
    cout << dp[steps - 1][m][n][(1 << n) - 1] << '\n';
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

5. Python Solution with Comments

```python
import sys
sys.setrecursionlimit(10**7)

def xor_fwt(a, inv=False):
    """In-place Walsh–Hadamard transform for XOR-convolution."""
    n = len(a)
    h = 1
    while h < n:
        for i in range(0, n, 2*h):
            for j in range(i, i+h):
                x = a[j]
                y = a[j+h]
                a[j]   = x + y
                a[j+h] = x - y
        h <<= 1
    if inv:
        # divide each element by n
        for i in range(n):
            a[i] //= n

def main():
    data = sys.stdin.read().split()
    it = iter(data)
    N = int(next(it))
    M = int(next(it)) - 1  # zero-based
    R = [list(map(int, (next(it) for _ in range(N)))) for __ in range(N)]

    # best_size[k] = minimal rounds to handle k players = ceil(log2(k))
    best_size = [0]*(N+1)
    for k in range(1, N+1):
        best_size[k] = best_size[k>>1] + 1
    STEPS = best_size[N-1] + 1

    FULL = (1<<N) - 1
    # dp[s][w][k] = FWT-domain vector of length 2^N
    dp = [[[[] for _ in range(N+1)] for __ in range(N)] for ___ in range(STEPS)]

    # Base: single-player subtrees at round 0
    for i in range(N):
        vec = [0]*(1<<N)
        vec[1<<i] = 1
        xor_fwt(vec, inv=False)
        dp[0][i][1] = vec

    # Precompute splits of size k into a+b
    splits = [(a,b) for a in range(1,N+1) for b in range(1,N+1) if a+b<=N]
    splits.sort(key=lambda x: max(x))

    # Merge subtrees
    for a,b in splits:
        k = a + b
        need = best_size[k-1] + 1
        for s1 in range(best_size[a-1], need):
            for s2 in range(best_size[b-1], need):
                if max(s1, s2) + 1 != need:
                    continue
                for x in range(N):
                    VA = dp[s1][x][a]
                    if not VA: continue
                    for y in range(N):
                        if x == y: continue
                        VB = dp[s2][y][b]
                        if not VB: continue
                        # Determine winner when x meets y
                        w = x if R[x][y] else y
                        if not dp[need-1][w][k]:
                            dp[need-1][w][k] = [0]*(1<<N)
                        VC = dp[need-1][w][k]
                        # pointwise multiply-add
                        for i in range(1<<N):
                            VC[i] += VA[i] * VB[i]

    # Extract answer
    res = dp[STEPS-1][M][N]
    if not res:
        print(0)
        return
    xor_fwt(res, inv=True)
    print(res[FULL])

if __name__ == "__main__":
    main()
```
