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 
const int m=1e9+7;
int n;
void matmul(int a[51][51], int b[51][51])
{
    int c[51][51]={};
    for(int i=0; i<n; ++i)
    {
        for(int j=0; j<n; ++j)
        {
            for(int k=0; k<n; ++k)
            {
                c[i][j]=(c[i][j] + a[i][k]*b[k][j]%m )%m;
            }
        }
    }
    for(int i=0; i<n; ++i)
    {
        for(int j=0; j<n; ++j)
        {
            a[i][j]=c[i][j];
        }
    }
}
void cpy(int a[51][51], int b[51][51])
{
    for(int i=0; i<n; ++i)
    {
        for(int j=0; j<n; ++j)
            a[i][j]=b[i][j];
    }
}
void powr(int r, int a[51][51], int ans[51][51])
{
    if(r==1)
    {
        cpy(ans, a);
        return;
    }
    powr(r>>1, a, ans);
    matmul(ans, ans);
    if(r&1)
    {
        matmul(ans, a);
    }
}
int32_t main()
{
    int k;
    cin>>n>>k;
    int a[51][51];
    for(int i=0; i<n; ++i)
    {
        for(int j=0; j<n; ++j)
        {
            cin>>a[i][j];
        }
    }
    int res[51][51]={};
    powr( k, a,  res);
    int ans=0;
    for(int i=0; i<n; ++i)
    {
        for(int j=0; j<n; ++j)
        {
            ans=(ans+res[i][j])%m;
        }
    }
    cout<<ans;
}