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 ll long long
#define fr(i,j,k) for(int i=j;i<k;i++)
#define f(n) fr(i,0,n)
#define f1(n) fr(i,1,n+1)
#define pb push_back
#define F first
#define S second
#define all(x) x.begin(), x.end()
const int mod = 1e9 + 7;
const int maxn = 100005;
struct mat {
    ll a[100][100];
    int n;
    mat() {
        memset(a, 0, sizeof(a));
    }
    mat operator * ( const mat &b ) const {
        mat ret;
        ret.n =b.n;
        for( int i = 0; i < n; ++i )
            for( int j = 0; j < n; ++j )
                for( int k = 0; k < n; ++k )
                    ret.a[i][j] = ( ret.a[i][j] + a[i][k] * b.a[k][j] ) % mod;
        return ret;
    }
};
ll solve(ll n, ll k) {
	mat now;
	now.n = n;
	f(n) {
		fr(j,0,n) {
			cin >> now.a[i][j];
		}
	}
	mat ans;
	ans.n = n;
	f(n) {
		ans.a[i][0] = 1;
	}
	while (k) {
		if (k & 1) {
			ans = now * ans;
		}
		now = now * now;
		k >>= 1;
	}
	ll sum = 0;
	f(n) {
		sum += ans.a[i][0];
		sum %= mod;
	}
	return sum;
}
void go() {
	ll n, m, k;
	cin >> n >> k;
	cout << solve(n, k) << '\n';

}
int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int c = 0;
    int t;
    if (!c) {
        t = 1;
    }
    else {
        cin >> t;
    }
    while (t--) {
        go();
    }
    
}