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>
using namespace std;
typedef long long ll;
typedef pair<int, int> pi;
typedef vector<int> vi;



typedef vector<vector<ll>> matrix;

const ll mod = 1e9 + 7;

matrix matmult(matrix a, matrix& b) {
    int p = a.size(), q = a[0].size(), r = b[0].size();
    matrix c(p, vector<ll>(r, 0));
    for (int i = 0; i < p; ++i) {
        for (int j = 0; j < r; ++j) {
            for (int k = 0; k < q; ++k) {
                c[i][j] += a[i][k] * b[k][j];
                c[i][j] %= mod;
            }
        }
    }
    return c;
}

matrix matpow(matrix& a, ll b) {
    matrix res(a.size(), vector<ll>(a.size(), 0));
    for (int i = 0; i < a.size(); ++i) res[i][i] = 1;
    while (b) {
        if (b & 1) res = matmult(res, a);
        a = matmult(a, a);
        b >>= 1;
    }
    return res;
}

void Main() {
    ll N, K;
    cin >> N >> K;
    matrix A(N, vector<ll>(N));
    for (int i = 0; i < N; ++i) {
        for (int j = 0; j < N; ++j) {
            cin >> A[i][j];
        }
    }
    A = matpow(A, K);
    ll ans = 0;
    for (int i = 0; i < N; ++i) {
        for (int j = 0; j < N; ++j) {
            ans += A[i][j];
            ans %= mod;
        }
    }
    cout << ans << '\n';
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    #ifdef _DEBUG
//    freopen("in.txt", "r", stdin);
//    freopen("out.txt", "w", stdout);
    #endif
    Main();
}