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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
import java.io.*;
class Main {
    static long MOD = (long)(1e9+7);
    static long[][] multiply(long[][] a, long[][] b) {
        int n = a.length;
        long[][] res = new long[n][n];
        for(int i = 0; i<n; i++) {
            for(int j = 0; j<n; j++) {
                for(int k = 0; k<n; k++) {
                    res[i][k] = (res[i][k] + (a[i][j]%MOD * 1l * b[j][k]%MOD)%MOD)%MOD;
                }
            }
        }
        return res;
    }
    public static void main (String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        long k = sc.nextLong();
        long[][] graph = new long[n][n];
        for(int i = 0; i<n; i++)
            for(int j = 0; j<n; j++)
                graph[i][j] = sc.nextLong();
        long[][] res = new long[n][n];
        for(int i = 0; i<n; i++)
            res[i][i] = 1l;
        while(k>0) {
            if(k%2==1)
                res = multiply(res, graph);
            graph = multiply(graph, graph);
            k/=2;
        }
        long ans = 0;
        for(int i = 0; i<n; i++) {
            for(int j = 0; j<n; j++)
                ans = (ans+res[i][j])%MOD;
        }
        System.out.println(ans);
    }
}