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<bits/stdc++.h>
using namespace std;
#define int long long
#define pb push_back
#define myPair pair<int,int>
#define mp make_pair
#define fi first
#define se second
#define qmax priority_queue<int>
#define qmin priority_queue<int,vector<int>,greater<int>>
#define all(x) x.begin(),x.end()
#define FOR(i,a,b) for(int i = a; i <= b; i++)
#define FORR(i,a,b) for(int i = a; i >= b; i--)
#define sz(a) (int)a.size()

vector<vector<int>> I;
int n, k;
const int MOD = 1e9 + 7;

vector<vector<int>> get() {
	vector<vector<int>> ans(n, vector<int> (n, 0));
	for(int i = 0; i < n; i++) {
		ans[i][i] = 1;
	}
	return ans;
}

vector<vector<int>> multiply(vector<vector<int>> A, vector<vector<int>> B) {
	vector<vector<int>> ans(n, vector<int> (n, 0));
	for(int i = 0; i < n; i++) {
		for(int j = 0; j < n; j++) {
			for(int k = 0; k < n; k++) {
				ans[i][j] = (ans[i][j] + (A[i][k] * B[k][j]) % MOD) % MOD;
			}
		}
	}
	return ans;
}

vector<vector<int>> power(vector<vector<int>> A, int k) {
	if(k == 0)
		return I;
	if(k % 2) {
		auto ans = power(A, (k - 1) / 2);
		ans = multiply(ans, ans);
		return multiply(ans, A);
	}
	else {
		auto ans = power(A, k / 2);
		return multiply(ans, ans);
	}
}

int32_t main()
{
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif

	ios_base::sync_with_stdio(false);
	cin.tie(NULL);

	cin >> n >> k;
	I = get();
	vector<vector<int>> g(n, vector<int> (n));
	for(auto &x : g) {
		for(int &v : x) {
			cin >> v;
		}
	}
	auto ans = power(g, k);
	int res = 0;
	for(auto x : ans) {
		for(int v : x) {
			res = (res + v) % MOD;
		}
	}
	cout << res;

	return 0;
}