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: Wrong Answer)

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

//vector<vector<int> > ;
vector< vector<ll int> > multiply(vector< vector<ll int> > a, vector< vector<ll int> > b)
{
    ll int m=a.size();
    ll int n=a[0].size();
//cout<<" hello1\n";
vector< vector<ll int> >v (m, vector<ll int> (n));
for(int i=0;i<m;i++)
{for(int j=0;j<n;j++)
{ll int sum=0;
    for(int k=0;k<m;k++)
    {sum=(sum+((a[k][i]%mod)   * (b[j][k]%mod) )%mod )%mod;

    }
v[i][j]=sum;

    }


}
//cout<<"hello\n";
return v;
}
vector< vector<ll int> > power(vector< vector<ll int> > a,ll int p)
{if(p==1)
    return a;
  if(p&1 == 1)
  {
    return multiply(a,power(a,p-1) );
  }
  else
  {vector< vector<ll int> > c=power(a,p/2);
    return multiply(c,c);

  }

}

int main()
{
   
   ll int n,k; 
    cin>>n>>k;
    vector< vector<ll int> > v(n,vector<ll int> (n));
    for(int i=0;i<n;i++)
    {for(int j=0;j<n;j++)
    cin>>v[i][j];
        
    }
    
    vector< vector<ll int> > v1=power(v,k);
ll int sum=0;
for(int i=0;i<n;i++)
{for(int j=0;j<n;j++)
{sum=(sum%mod + v1[i][j]%mod)%mod;
    
}
}


   cout<<sum<<endl;

}