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 <iostream>
#include <vector>
using namespace std;

typedef long long ll;
typedef vector<ll> vl;

const ll MOD = (ll)1e9 + 7;
int N;
ll K;
vector<vl> A, ans;

void input_init(void) {
	cin >> N >> K;
	A.resize(N, vl(N));
	for (int j = 0; j < N; ++j) {
		for (int i = 0; i < N; ++i)
			cin >> A[j][i];
	}
	ans.resize(N, vl(N, 0));
	for (int i = 0; i < N; ++i)
		ans[i][i] = (ll)1;
}

void pow_A(ll x) {
	while (x > (ll)0) {
		if ((x & (ll)1) == (ll)1) {
			vector<vl> tmp(N, vl(N, (ll)0));
			for (int j = 0; j < N; ++j) {
				for (int k = 0; k < N; ++k) {
					for (int i = 0; i < N; ++i) {
						tmp[j][i] += (ans[j][k] * A[k][i]) % MOD;
						tmp[j][i] %= MOD;
					}
				}
			}
			for (int j = 0; j < N; ++j) {
				for (int i = 0; i < N; ++i)
					ans[j][i] = tmp[j][i];
			}
		}
		vector<vl> tmp(N, vl(N, (ll)0));
		for (int j = 0; j < N; ++j) {
			for (int k = 0; k < N; ++k) {
				for (int i = 0; i < N; ++i) {
					tmp[j][i] += (A[j][k] * A[k][i]) % MOD;
					tmp[j][i] %= MOD;
				}
			}
		}
		for (int j = 0; j < N; ++j) {
			for (int i = 0; i < N; ++i)
				A[j][i] = tmp[j][i];
		}
		x >>= 1;
	}
}

ll solve(void) {
	ll ret = 0;
	for (int j = 0; j < N; ++j) {
		for (int i = 0; i < N; ++i) {
			ret += ans[j][i] % MOD;
			ret %= MOD;
		}
	}
	return ret;
}

int main(void) {
	cin.tie(0);
	ios::sync_with_stdio(false);

	input_init();
	pow_A(K);
	ll val = solve();

	cout << val << '\n';
	return 0;
}