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 (GCC 9.2.1) (judge verdict: Accepted)

#include <stdio.h>

const int Mod = 1000000007;

void prod_matrix(int N, long long a[][51], long long b[][51], long long c[][51])
{
	int i, j, k;
	for (i = 1; i <= N; i++) {
		for (j = 1; j <= N; j++) {
			for (k = 1, c[i][j] = 0; k <= N; k++) c[i][j] = (c[i][j] + a[i][k] * b[k][j]) % Mod;
		}
	}
}

void pow_matrix(int N, long long a[][51], long long K, long long b[][51])
{
	int i, j;
	long long c[51][51], d[51][51];
	for (i = 1; i <= N; i++) {
		for (j = 1; j <= N; j++) {
			b[i][j] = (i == j)? 1: 0;
			c[i][j] = a[i][j];
		}
	}
	for (; K > 0; K >>= 1) {
		if (K % 2 == 1) {
			prod_matrix(N, b, c, d);
			for (i = 1; i <= N; i++) for (j = 1; j <= N; j++) b[i][j] = d[i][j];
		}
		prod_matrix(N, c, c, d);
		for (i = 1; i <= N; i++) for (j = 1; j <= N; j++) c[i][j] = d[i][j];
	}
}

int main()
{
	int i, j, N;
	long long K, a[51][51];
	scanf("%d %lld", &N, &K);
	for (i = 1; i <= N; i++) for (j = 1; j <= N; j++) scanf("%lld", &(a[i][j]));
	
	long long b[51][51], ans = 0;
	pow_matrix(N, a, K, b);
	for (i = 1; i <= N; i++) for (j = 1; j <= N; j++) ans += b[i][j];
	printf("%lld\n", ans % Mod);
	fflush(stdout);
	return 0;
}