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>
typedef long long int ll;
typedef long double ld;
#define pb push_back
#define pii pair < ll , ll >
#define F first
#define S second
#define int long long int
#define sync ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define err(A) cout << #A << " = " << (A) << endl
using namespace std;
const int N=64,mod=1e9+7;

ll dp[N][N][N];
ll d[N][N][N];
ll a[N][N];
int32_t main(){
    ll n,k;
    cin >> n >> k;
    for (int i=1;i<=n;i++){
        for (int j=1;j<=n;j++){
            cin >> a[i][j];
            if (a[i][j]){
                dp[i][j][0]=1;
            }
        }
    }
    vector <int> b;
    while(k>0){
        b.pb(k%2);
        k/=2;
    }
    for (int m=1;m<=b.size();m++){
        for (int i=1;i<=n;i++){
            d[i][i][b.size()]=1;
            for (int j=1;j<=n;j++){

                /// dp[i][j][m];
                for (int l=1;l<=n;l++){
                    dp[i][j][m]+=dp[i][l][m-1]*dp[l][j][m-1]%mod;
                    dp[i][j][m]%=mod;
                }
                //cout << i << " " << j << " " << m << " " << dp[i][j][m] << endl;
            }
        }
    }
    for (int m=b.size()-1;m>-1;m--){
        for (int i=1;i<=n;i++){
                for (int j=1;j<=n;j++){
                    if (b[m]==0){
                        d[i][j][m]=d[i][j][m+1];
                       // cout << 2 << endl;
                        continue;
                    }
                    for (int l=1;l<=n;l++){
                        d[i][j][m]+=dp[i][l][m]*d[l][j][m+1]%mod;
                        d[i][j][m]%=mod;
                    }
                    //cout << i << " " << j << " " << m << " " << d[i][j][m] << endl;

                }
        }
    }
    ll ans=0;
    for (int i=1;i<=n;i++){
        for (int j=1;j<=n;j++){
            ans+=d[i][j][0];
            ans%=mod;
        }
    }
    cout << ans;
}