<|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.  
   - Let STEPS = best_size[N] be the total number of rounds (equals T).  
   - We define dp[s][w][k] as a vector of length 2ᴺ in FWT‐domain, where:  
       • s = number of rounds used so far (0≤s≤STEPS)  
       • w = winner of this subtree (0≤w<N)  
       • k = the number of players in the subtree (1≤k≤N)  
     and dp[s][w][k][mask] (in ordinary domain) = number of ways to choose exactly the set of players = mask (popcount(mask)=k) in a subtree of height s that produces winner w, 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_fwt.  

c. Transitions (merging two subtrees)  
   - Consider splitting a subtree of size k into sizes a and b with a+b=k.  
   - Let s₁ and s₂ be rounds used by the left and right subtrees respectively; to form the parent subtree we need s = max(s₁, s₂)+1. We only allow s₁≥best_size[a], s₂≥best_size[b], and s = best_size[k].  
   - For every possible left‐winner x and right‐winner y, we know the final winner w = x if R[x][y]=1 else y.  
   - In FWT‐domain we simply do:  
       dp[s][w][k] += dp[s₁][x][a]  *  dp[s₂][y][b]   (elementwise on the 2ᴺ‐vectors)  
     accumulation is a pointwise multiply‐and‐add across all 2ᴺ positions.  

d. Final answer  
   - After filling dp, look at dp[STEPS][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 STEPS ≤ 4 (since N≤16, T=⌈log₂16⌉=4).  
   - Number of dp entries: O(STEPS·N·N). Each is a length 2ᴺ vector.  
   - Each FWT or inverse FWT costs O(N·2ᴺ).  
   - Each merge operation is a pointwise multiply of two 2ᴺ‐vectors, O(2ᴺ).  
   - Total is a few hundred passes over vectors of length ≤65 536, which fits in 1 s in optimized C++.

4. C++ implementation with detailed comments  
```cpp
#include <bits/stdc++.h>
using namespace std;

/*
 Fast Walsh-Hadamard Transform (in-place) for XOR-convolution.
 If inv==true, perform inverse transform (divide each element by length at end).
*/
void xor_fwt(vector<uint64_t> &a, bool inv=false) {
    int n = a.size();
    for(int len = 1; len < n; len <<= 1) {
        for(int i = 0; i < n; i += 2*len) {
            for(int j = 0; j < len; j++) {
                uint64_t u = a[i+j];
                uint64_t v = a[i+j+len];
                a[i+j]     = u + v;
                a[i+j+len] = u - v;
            }
        }
    }
    if(inv) {
        // divide every element by n
        for(auto &x: a) x /= n;
    }
}

// Elementwise multiply-and-add: res += A * B  (all same length)
void multiply_add(vector<uint64_t> &res,
                  const vector<uint64_t> &A,
                  const vector<uint64_t> &B) {
    size_t n = res.size();
    for(size_t i = 0; i < n; i++) {
        res[i] += A[i] * B[i];
    }
}

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

    int N, M;
    cin >> N >> M;
    M--;  // zero-based friend index

    vector<vector<int>> R(N, vector<int>(N));
    for(int i = 0; i < N; i++)
        for(int j = 0; j < N; j++)
            cin >> R[i][j];

    // Precompute minimal rounds needed for k players: best_size[k] = ceil(log2(k))
    vector<int> best_size(N+1, 0);
    for(int k = 1; k <= N; k++) {
        best_size[k] = best_size[k>>1] + 1;
    }
    int STEPS = best_size[N];  // total rounds

    int FULL = (1<<N) - 1;
    // dp[s][w][k] is a vector<uint64_t> in FWT domain, length=2^N
    // Use a 3D container of size [STEPS+1][N][N+1]
    vector<vector<vector<vector<uint64_t>>>> dp(
        STEPS+1,
        vector<vector<vector<uint64_t>>>(N, vector<vector<uint64_t>>(N+1))
    );

    // Base case: single-player subtrees at round 0
    for(int i = 0; i < N; i++) {
        dp[0][i][1].assign(1<<N, 0);
        dp[0][i][1][1<<i] = 1;      // mask with only player i
        xor_fwt(dp[0][i][1], false);
    }

    // Precompute all ways to split a subtree of size k into (a, b)
    vector<pair<int,int>> splits;
    for(int a = 1; a <= N; a++){
        for(int b = 1; b + a <= N; b++){
            splits.emplace_back(a, b);
        }
    }
    // Sort splits by max(a,b) so we build smaller subtrees first
    sort(splits.begin(), splits.end(),
         [&](auto &A, auto &B){
             return max(A.first, A.second) < max(B.first, B.second);
         });

    // DP transitions: merge two subtrees of sizes a, b into size a+b
    for(auto &pr : splits) {
        int a = pr.first, b = pr.second, k = a+b;
        int need_round = best_size[k];  // final round for size k
        // enumerate all ways left uses s1 rounds, right uses s2 rounds
        for(int s1 = best_size[a]; s1 < need_round; s1++){
            for(int s2 = best_size[b]; s2 < need_round; s2++){
                if(max(s1, s2) + 1 != need_round) continue;
                // for each possible winners x,y
                for(int x = 0; x < N; x++){
                    auto &VA = dp[s1][x][a];
                    if(VA.empty()) continue;
                    for(int y = 0; y < N; y++){
                        if(x == y) continue;
                        auto &VB = dp[s2][y][b];
                        if(VB.empty()) continue;
                        int w = R[x][y] ? x : y;  // final winner
                        auto &VC = dp[need_round][w][k];
                        if(VC.empty()) {
                            VC.assign(1<<N, 0);
                        }
                        multiply_add(VC, VA, VB);
                    }
                }
            }
        }
    }

    // Fetch the result for full subtree of size N with friend M as winner
    auto &ANS_FWT = dp[STEPS][M][N];
    if(ANS_FWT.empty()) {
        cout << 0 << "\n";
        return 0;
    }
    // Inverse FWT back to normal domain
    xor_fwt(ANS_FWT, true);
    // Answer is the coefficient for mask = FULL
    cout << ANS_FWT[FULL] << "\n";
    return 0;
}
```

5. Python implementation with detailed 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]

    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+1)]

    # 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]
        for s1 in range(best_size[a], need):
            for s2 in range(best_size[b], 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][w][k]:
                            dp[need][w][k] = [0]*(1<<N)
                        VC = dp[need][w][k]
                        # pointwise multiply-add
                        for i in range(1<<N):
                            VC[i] += VA[i] * VB[i]

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

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