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;
#define int long long
typedef long long ll;

const int mod = 1e9+7;

vector<vector<ll> > mult(vector<vector<ll> > a, vector<vector<ll> > b) {
    if(a.size() == 0) return {};
    if(a[0].size() == 0) return {};
    if(b.size() == 0) return {};
    if(b[0].size() == 0) return {};
    if(a[0].size() != b.size()) return {};
    int resultRow = a.size(), resultCol = b[0].size(), n = a[0].size();
    vector<vector<ll> > product(resultRow, vector<ll>(resultCol));
    for(int i = 0; i < resultRow; ++i) {
        for(int j = 0; j < resultCol; ++j) {
            product[i][j] = 0;
            for(int k = 0; k < n; ++k) {
                product[i][j] += a[i][k] * b[k][j];
                product[i][j] %= mod;
            }
        }
    }
    return product;
}

vector<vector<ll> > power(vector<vector<ll> > matrix, ll b) {
    if (b <= 1) return matrix;
    vector<vector<ll> > temp = power(matrix, b/2);
    if (b % 2 == 0) return mult(temp, temp);
    return mult(mult(temp, temp), matrix);
}

int32_t main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int n,k;
    cin >> n >> k;
    vector<vector<int>> arr(n, vector<int>(n));
    for(int i = 0; i < n; ++i) {
        for(int j = 0; j < n; ++j) {
            cin >> arr[i][j];
        }
    }
    vector<vector<int>> vect(n, vector<int>(1,1));
    vect = mult(power(arr, k), vect);
    int res = 0;
    for(int i = 0; i < n; ++i) {
        res += vect[i][0];
        res %= mod;
    }
    cout << res << '\n';
    return 0;
}