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 <iostream>
#include <string>
#include <map>
#include <set>
#include <cstring>
#include <vector>
#include <cmath>
#include <algorithm>

using namespace std;
typedef long long ll;
const ll mod = 1e9 + 7;

//Matrix exponentiation
struct Matrix {
	int n; vector<vector<ll>> v; Matrix() {}
	Matrix(vector<vector<ll>> &_v) : n(_v.size()) { v = move(_v); }
	Matrix(int _n, ll _x) : n(_n), v(n, vector<ll>(n, _x)) {}
	Matrix(int _n) : n(_n), v(n, vector<ll>(n, 0)) {
		for (int i = 0; i < _n; i++) { v[i][i] = 1; }
	}
	vector<ll>& operator[](int idx) { return v[idx]; }

	Matrix operator *(Matrix &rhs) {
		Matrix res(n, 0);
		for (int i = 0; i < n; i++)
			for (int j = 0; j < n; j++)
				for (int k = 0; k < n; k++)
					res[i][j] = (res[i][j] + v[i][k] * rhs[k][j]) % mod;
		return move(res);
	}

	Matrix operator ^(ll x) {
		Matrix res(n), a(v);
		for (; x; x >>= 1, a = a * a)
			if (x & 1) { res = res * a; }
		return move(res);
	}
};

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);

	ll n, k; cin >> n >> k;
	vector<vector<ll>> v(n, vector<ll>(n, 0));
	for (int i = 0; i < n; i++)
		for (int j = 0; j < n; j++)
			cin >> v[i][j];

	Matrix M(v);
	M = M ^ (k);
	
	ll ans = 0;
	for (int i = 0; i < n; i++)
		for (int j = 0; j < n; j++)
			ans = (ans + M[i][j]) % mod;

	cout << ans << '\n';
}