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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<iostream>
#include<bits/stdc++.h>
#define int long long
using namespace std;
#define mod 1000000007


vector<vector<int>> multiply(vector<vector<int>> A, vector<vector<int>> B)
{
	int sz = (int)A.size();

	vector<vector<int>> C(sz , vector<int>(sz,0));
	for(int i=0;i<sz;i++)
		for(int j=0;j<sz;j++)
			for(int x=0;x<sz;x++)
				C[i][j] = ( C[i][j] + A[i][x]*B[x][j] )%mod;

	return C;
}



vector<vector<int>> pow(vector<vector<int> > &adj, int n){

	if(n==1)
		return adj;

	vector<vector<int>> res = pow(adj,n/2);
	if(n&1)
		return multiply(multiply(adj,res),res); 
	else
		return multiply(res,res);

}




int32_t main(){

#ifndef ONLINE_JUDGE
    // for getting input from input.txt
    freopen("input.txt", "r", stdin);
    // for writing output to output.txt
    freopen("output.txt", "w", stdout);
#endif

    ios_base::sync_with_stdio(false);
        cin.tie(NULL);

	int n,k;
	cin>>n>>k;
	vector<vector<int>> adj(n,vector<int>(n,0));
	for(int i=0;i<n;i++)
		for(int j=0;j<n;j++)
			cin>>adj[i][j];


	vector<vector<int>> a = pow(adj,k);

	int ans = 0;
	for(int i=0;i<n;i++)
	{
		for(int j=0;j<n;j++)
		{
			ans = (ans+a[i][j])%mod;
		}
	}

	cout<<ans<<endl;



}