Problem:
Score : 100 points

Problem Statement

There is a simple directed graph G with N vertices, numbered 1, 2, \ldots, N.

For each i and j (1 \leq i, j \leq N), you are given an integer a_{i, j} that represents whether there is a directed edge from Vertex i to j.
If a_{i, j} = 1, there is a directed edge from Vertex i to j; if a_{i, j} = 0, there is not.

Find the number of different directed paths of length K in G, modulo 10^9 + 7.
We will also count a path that traverses the same edge multiple times.

Constraints

All values in input are integers.

1 \leq N \leq 50

1 \leq K \leq 10^{18}

a_{i, j} is 0 or 1.

a_{i, i} = 0

Input

Input is given from Standard Input in the following format:

N K
a_{1, 1} \ldots a_{1, N}
:
a_{N, 1} \ldots a_{N, N}

Output

Print the number of different directed paths of length K in G, modulo 10^9 + 7.

Sample Input 1

4 2
0 1 0 0
0 0 1 1
0 0 0 1
1 0 0 0

Sample Output 1

6

G is drawn in the figure below:

There are six directed paths of length 2:

1 → 2 → 3

1 → 2 → 4

2 → 3 → 4

2 → 4 → 1

3 → 4 → 1

4 → 1 → 2

Sample Input 2

3 3
0 1 0
1 0 1
0 0 0

Sample Output 2

3

G is drawn in the figure below:

There are three directed paths of length 3:

1 → 2 → 1 → 2

2 → 1 → 2 → 1

2 → 1 → 2 → 3

Sample Input 3

6 2
0 0 0 0 0 0
0 0 1 0 0 0
0 0 0 0 0 0
0 0 0 0 1 0
0 0 0 0 0 1
0 0 0 0 0 0

Sample Output 3

1

G is drawn in the figure below:

There is one directed path of length 2:

4 → 5 → 6

Sample Input 4

1 1
0

Sample Output 4

0

Sample Input 5

10 1000000000000000000
0 0 1 1 0 0 0 1 1 0
0 0 0 0 0 1 1 1 0 0
0 1 0 0 0 1 0 1 0 1
1 1 1 0 1 1 0 1 1 0
0 1 1 1 0 1 0 1 1 1
0 0 0 1 0 0 1 0 1 0
0 0 0 1 1 0 0 1 0 1
1 0 0 0 1 0 1 0 0 0
0 0 0 0 0 1 0 0 0 0
1 0 1 1 1 0 1 1 1 0

Sample Output 5

957538352

Be sure to print the count modulo 10^9 + 7.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <cstdio>
typedef long long lld;
const int MAXN = 55;
const int MOD = 1e9+7;
lld K;
int N;

struct Matrix {
  int data[MAXN][MAXN];

  int& at(int i, int j) { return data[i][j]; }

  Matrix& operator *= (Matrix other) {
    Matrix copy = *this;
    for (int i = 0; i < N; ++i) {
      for (int j = 0; j < N; ++j) {
        data[i][j] = 0;
        for (int k = 0; k < N; ++k) {
          int val = 1LL * copy.at(i, k) * other.at(k, j) % MOD;
          data[i][j] = (1LL * val + data[i][j]) % MOD;
        }
      }
    }
    return *this;
  }

  void identity() {
    for (int i = 0; i < N; ++i) {
      for (int j = 0; j < N; ++j) {
        data[i][j] = (i == j) ? 1 : 0;
      }
    }
  }

  Matrix pow(lld exp) {
    Matrix ret;
    ret.identity();
    while (exp) {
      if (exp & 1LL) {
        ret *= *this;
      }
      *this *= *this;
      exp /= 2;
    }
    return ret;
  }

  int sum() const {
    int ans = 0;
    for (int i = 0; i < N; ++i) {
      for (int j = 0; j < N; ++j) {
        ans = (1LL * data[i][j] + ans) % MOD;
      }
    }
    return ans;
  }
} A;

int main() {
  scanf(" %d%lld", &N, &K);
  for (int i = 0; i < N; ++i) {
    for (int j = 0; j < N; ++j) {
      scanf("%d", &A.at(i, j));
    }
  }
  printf("%d\n", A.pow(K).sum());
  return 0;
}