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 <bits/stdc++.h>
#define int long long
using namespace std;

int MOD = 1e9 + 7;

struct Matrix {

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

        Matrix(vector < vector < int > > M) : mat(M), n(M.size()), m(M[0].size()) {}

        static Matrix id(int n) {
                vector < vector < int > > I(n, vector <int> (n, 0));
                for(int i = 0; i < n; i++) {
                        I[i][i] = 1;
                }
                return Matrix(I);
        }

        Matrix operator * (Matrix A) {
                int N = n;
                int M = A.m;
                int O = m;
                vector < vector < int > > Temp(N, vector <int> (M, 0));

                for(int i = 0; i < N; i++) {
                        for(int j = 0; j < M; j++) {
                                for(int k = 0; k < O; k++) {
                                        Temp[i][j] = (Temp[i][j] + mat[i][k] * A.mat[k][j]) % MOD;
                                }
                        }
                }
                return Matrix(Temp);
        }

        friend ostream& operator << (ostream &out, Matrix A) ;


};

ostream& operator << (ostream &out, Matrix A) {
        for(int i = 0; i < A.n; i++) {
                if(i) cout << '\n';
                for(int j = 0; j < A.m; j++) {
                        cout << A.mat[i][j] << " ";
                }
        }
}

Matrix BigMod(int n, Matrix M) {
        Matrix ret = M.id(M.n);
        while(n) {
                if(n & 1) {
                        ret = ret * M;
                }
                n /= 2;
                M = M * M;

        }
        return ret;
}

int32_t main() {
        int n, k; cin >> n >> k;
        vector <vector <int> > r(n, vector <int> (n, 0));
        for(int i = 0; i < n; i++)
            for(int j = 0; j < n; j++)
                cin >> r[i][j];
        Matrix M(r);

        M = BigMod(k, M);

        int ans = 0;
        for(int i = 0; i < n; i++)
            for(int j = 0; j < n; j++)
                ans = (ans + M.mat[i][j]) % MOD;
        cout << ans << endl;
}