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 <iostream>
#include <iomanip>
#include <bits/stdc++.h>
#define ff first
#define ss second
#define M 1000000007
typedef long long int ll;
using namespace std;
ll power(ll a,ll b){
    ll ans = 1;
    while(b>0){
        if(b%2!=0){
            ans = ((ans)*(a));
        }
        a = ((a)*(a));
        b=b/2;
    }
    return ans;
}
ll add(ll a,ll b){
    ll temp = (a%M + b%M)%M;
    if(temp>M){
        temp=temp-M;
    }
    return temp;
}
vector <vector <ll>> mul(const vector <vector <ll>> &a,const vector <vector <ll>> &b){
    ll n = a.size();
    vector <vector <ll>> res(n,vector <ll> (n,0));
    for(ll i=0;i<n;i++){
        for(ll j=0;j<n;j++){
            for(ll k=0;k<n;k++){
                res[i][j]+=((a[i][k]%M)*(b[k][j]%M))%M;
                while(res[i][j]>M){
                    res[i][j]=res[i][j]%M;
                }
            }
        }
    }
    return res;
}
vector <vector <ll>> power(vector <vector <ll>> &a,ll k){
    ll n = a.size();
    vector <vector <ll>> res(n,vector <ll> (n,0));
    for(ll i=0;i<n;i++){
        res[i][i]=1;
    }
    while(k){
        if(k%2){
            res = mul(res,a);
        }
        a = mul(a,a);
        k=k/2;
    }
    return res;
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll n,k;
    cin>>n>>k;
    vector <vector <ll>> a(n,vector <ll> (n));
    for(ll i=0;i<n;i++){
        for(ll j=0;j<n;j++){
            cin>>a[i][j];
        }
    }
    vector <vector <ll>> res=power(a,k);
    ll ans = 0;
    for(ll i=0;i<n;i++){
        for(ll j=0;j<n;j++){
            ans=add(ans,res[i][j]);
        }
    }
    cout<<ans<<endl;
    return 0;
}