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: Runtime Error)

#include<bits/stdc++.h>
using namespace std;
int n ;
int tree[51][51];

vector< vector< int > > memo(51, vector<int> (200,-1)); 

int find(int node, int k)
{   
    
    //cout << " node " << node << " k " << k << endl; 
	if( k == 0 ) return 1 ; 
	if( memo[node][k] != -1 ) 
	   return memo[node][k]; 
    
    int res = 0 ; 
	for(int child = 0 ; child < n ; child ++)
	{
		if(tree[node][child])
		{
            res += find(child,k-1); 
		}
	}

	return memo[node][k] = res; 
}


void solve()
{     
       
       int k;  
       cin >> n >> k ;
       for(int i = 0 ; i < n ; ++i)
       	for(int j = 0; j < n ; ++j)
       		cin >> tree[i][j]; 
       
       //cout << " 1 " << endl; 
       int res = 0 ;
       for(int i = 0; i < n ; i ++ )
       {
           res += find(i,k);
           //cout << res << " "; 
       }
       cout << res; 

}

int main()
{

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

	
	 int test = 0, t = 1;
	 if(test) cin >> t;
	 for(int i = 0 ; i < t ; i ++)
	 {
         solve(); 
	 }

	return 0;
}