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;
long long a[100][100],b[100][100],c[100][100],ans,o[100][100],mod=1e9+7,k,n;
void mul(long long x)
{
    if (x==1)
    {
        for (int i=0; i<n; i++)
        {
            for (int j=0; j<n; j++)
            {
                c[i][j]=o[i][j];
            }
        }
        return;
    }
    if (x>1) mul(x/2);
    for (int i=0; i<n; i++)
    {
        for (int j=0; j<n; j++)
        {
            a[i][j]=c[i][j];
            b[i][j]=c[i][j];
            c[i][j]=0;
        }
    }
    for (int i=0; i<n; i++)
    {
        for (int j=0; j<n; j++)
        {
            for (int k=0; k<n; k++)
            {
                c[i][j]+=(a[i][k]*b[k][j])%mod;
                c[i][j]%=mod;
            }
        }
    }
    if (x%2==1)
    {
        for (int i=0; i<n; i++)
        {
            for (int j=0; j<n; j++)
            {
                a[i][j]=c[i][j];
                b[i][j]=o[i][j];
                c[i][j]=0;
            }
        }
        for (int i=0; i<n; i++)
        {
            for (int j=0; j<n; j++)
            {
                for (int k=0; k<n; k++)
                {
                    c[i][j]+=(a[i][k]*b[k][j])%mod;
                    c[i][j]%=mod;
                }
            }
        }
    }
    return ;
}
long long sum;
int main()
{
    cin>>n>>k;
    for (int i=0;i<n;i++)
    {
        for (int j=0;j<n;j++) cin>>o[i][j];
    }
    mul(k);
    for(int i=0;i<n;i++)
    {
            for (int j=0;j<n;j++)
        {
            sum+=c[i][j]%mod;
           // cout<< c[i][j]<<" ";
            sum%=mod;
        }
       // cout<<endl;
    }

    cout<<sum;
    return 0;
}