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: Wrong Answer)

import java.util.*;
import java.io.*;
public class Main {
    static long mod=1000000007L;
    public static void main(String args[]) {
        Scanner input=new Scanner(System.in);
        int n=input.nextInt();
        long k=input.nextLong();
        long arr[][]=new long[n][n];
        for(int i=0;i<n;i++) {
            for(int j=0;j<n;j++) {
                arr[i][j]=input.nextLong();
            }
        }
        long res[][]=new long[n][n];
        power(arr, res, k);
        long sum=0;
        for(int i=0;i<n;i++) {
            for(int j=0;j<n;j++) {
                sum+=res[i][j];
                sum%=mod;
            }
        }
        System.out.println(sum);
    }
    public static void power(long arr[][], long res[][], long n) {
        if(n==1) {
            for(int i=0;i<arr.length;i++) {
                for(int j=0;j<arr.length;j++) {
                    res[i][j]=arr[i][j];
                }
            }
            return;
        }
        power(arr,res,n/2);
        multiply(arr,arr,res);
        if(n%2==1) {
            multiply(res,arr,res);
        }
    }
    
    static void multiply(long a[][], long b[][], long res[][])  { 
        long mul[][] = new long[a.length][a.length]; 
        for (int i = 0; i < a.length; i++) { 
            for (int j = 0; j < a.length; j++)  { 
                mul[i][j] = 0; 
                for (int k = 0; k < a.length; k++) {
                    mul[i][j] += a[i][k] * b[k][j]; 
                    mul[i][j]%=mod;
                }
            } 
        }
        for (int i = 0; i < a.length; i++) {
            for (int j = 0; j < a.length; j++) {
                res[i][j] = mul[i][j]; 
            }
        }
    } 
}