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)

#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize ("Ofast")
#pragma GCC target ("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
 
 
#include <iostream>
#include <iomanip>
#include <sstream>
#include <cstring>
#include <vector>
#include <deque>
#include <queue>
#include <set>
#include <map>
#include <valarray>
#include <iterator>
#include <functional>
#include <limits>
#include <algorithm>
#include <numeric>
#include <cmath>
#include <cassert>
#include <unordered_map>
#include <unordered_set>
#include <stack>
using namespace std;
#define IOS ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
 
typedef int64_t ll;
 
#define endl '\n'
#define int ll
ll mod=1e9+7;
ll mod1=1e9+5;
ll power(ll a,ll b)
{
    if(b==0) return 1;
    else if(b%2==0)
        return power((((a%mod)*(a%mod))%mod),b/2)%mod;
    else return ((a%mod)*(power((((a%mod)*(a%mod))%mod),b/2)%mod))%mod;
}
ll mat[60][60];
void mult(ll mat1[60][60],ll mat2[60][60])
{

    ll res[51][51];
     for (int i = 0; i <= 50; i++) 
    { 
        for (int j = 0; j <=50; j++) 
        { 
            res[i][j] = 0; 
            for (int k = 0; k <=50; k++) 
                res[i][j]=(res[i][j]%mod +(mat1[i][k]%mod * mat2[k][j]%mod)%mod)%mod; 
        } 
    }
    for(int i=0;i<=50;++i)
    {
        for(int j=0;j<=50;++j)
            mat[i][j]=res[i][j];
    }
}
void expo(ll k)
{
    if(k==1)
        return ;
    else if(k%2==0)
    {
            mult(mat,mat);
            expo(k/2);
    }
    else
    {
        ll temp[60][60];
        for(int i=0;i<=50;++i)
            for(int j=0;j<=50;++j)
                temp[i][j]=mat[i][j];
            mult(mat,mat);
            
            expo(k/2);   
            mult(temp,mat);
    }
}
int32_t main()
{
    IOS
     
    

    int n,k;
    cin>>n>>k;
    memset(mat,0,sizeof(mat));
    for(int i=1;i<=n;++i)
    {
        for(int j=1;j<=n;++j)
            cin>>mat[i][j];
    }
    expo(k);
    int ans=0;
    for(int i=1;i<=n;++i)
    {
        for(int j=1;j<=n;++j)
            ans+=mat[i][j],ans%=mod;
    }
    cout<<ans;

}