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 <vector>
using namespace std;

#define rep(i,o,n) for(auto i=o;i<n;i++)

typedef vector<vector<long long>> matrix;

const int P=1000000007;
int n;


matrix multiply(matrix& m1,matrix& m2){
	matrix ans(n,vector<long long>(n,0));

	rep(i,0,n){
		rep(j,0,n){
			rep(k,0,n){
				ans[i][j] = (ans[i][j]+ m1[i][k]*m2[k][j])%P;
			}
		}
	}

	return ans;
}

matrix binexp(matrix G,long long k){
	if(k==0){
		matrix I(n,vector<long long>(n,0));
		rep(i,0,n)
			I[i][i]=1;
		return I;
	}
	matrix sqr = multiply(G,G);
	matrix res = binexp(sqr,k>>1);
	if(k&1)
		return multiply(G,res);
	return res;
}

int main(){
	long long k;
	cin>>n>>k;
	matrix G(n,vector<long long>(n,0));
	
	rep(i,0,n)
		rep(j,0,n)
			cin>>G[i][j];

	matrix A = binexp(G,k);
	
	int sum = 0;
	rep(i,0,n)
		rep(j,0,n)
			sum = (sum+A[i][j])%P;

	cout<<sum<<endl;
}