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 <stdio.h>
#include <algorithm>
#include <string>
#include <vector>
#include <numeric>
#include<queue>

const int mod = 1000000007;
//const int mod = 998244353;

class Matrix {
	int N;
	long long **d;
public:
	Matrix(int N) :N(N) {
		d = new long long*[N];
		for (int i = 0; i < N; i++) {
			d[i] = new long long[N] {};
		}
	}
	void set(long long** o) {
		for (int i = 0; i < N; i++) {
			for (int j = 0; j < N; j++) {
				d[i][j] = o[i][j];
			}
		}
	}
	void identity() {
		for (int i = 0; i < N; i++)d[i][i] = 1;
	}

	~Matrix() {
		for (int i = 0; i < N; i++) {
			delete[] d[i];
		}
		delete[] d;
	}
	long long* operator[](int x) const {
		return d[x];
	}
	Matrix& operator*=(const Matrix& o) {
		Matrix temp(N);
		for (int i = 0; i < N; i++) {
			for (int j = 0; j < N; j++) {
				for (int k = 0; k < N; k++) {
					temp[i][j] += d[i][k] * o[k][j];
					temp[i][j] %= mod;
				}
			}
		}
		for (int i = 0; i < N; i++) {
			for (int j = 0; j < N; j++) {
				d[i][j] = temp[i][j];
			}
		}
		return *this;
	}
	long long* operator*(long long* v) {
		long long* res = new long long[N] {};
		for (int i = 0; i < N; i++) {
			for (int j = 0; j < N; j++) {
				res[i] += d[i][j] * v[j];
			}
		}
		return res;
	}
	void print() {
		for (int i = 0; i < N; i++) {
			for (int j = 0; j < N; j++) {
				printf("%lld ", d[i][j]);
			}
			printf("\n");
		}
	}
};

int main() {
	int N;
	long long K;
	scanf("%d %lld", &N, &K);
	long long **G = new long long*[N];
	for (int i = 0; i < N; i++) {
		G[i] = new long long[N];
	}
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < N; j++) {
			scanf("%lld", G[i] + j);
		}
	}
	Matrix g(N);
	g.set(G);
	Matrix r(N);
	r.identity();
	while (K != 0) {
		if (K & 1) {
			r *= g;
		}
		K >>= 1;
		g *= g;
	}
	long long res = 0;
	for (int i = 0; i < N; i++) for (int j = 0; j < N; j++) {
		res += r[i][j];
		res %= mod;
	}
	printf("%lld", res);
	return 0;
}