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 <bits/stdc++.h>
#define va first
#define vb second
#define lb lower_bound
#define pb push_back
#define all(v) v.begin(),v.end()
#define fio ios_base::sync_with_stdio(0);cin.tie(0);
using namespace std;
using ll = long long;
using lf = long double;
using pii = pair<int,int>;
using pll = pair<ll,ll>;
const ll mod = 1e9 + 7;

int n;
ll k;

inline void add(ll &x, ll y){
    x = (x + y) % mod;
}

struct mat{
    ll a[50][50];
    mat operator* (mat x){
        mat c;
        for(int i = 0; i < n; i++){
            for(int j = 0; j < n; j++){
                c.a[i][j] = 0;
                for(int k = 0; k < n; k++){
                    add(c.a[i][j], a[i][k] * x.a[k][j] % mod);
                }
            }
        }
        return c;
    }
    void show(string S = "Mat"){
        cout << S << '\n';
        for(int i = 0; i < n; i++){
            for(int j = 0; j < n; j++) cout << a[i][j] << ' ';
            cout << '\n';
        }
    }
};
mat A, T;
int main(){
    fio;
    cin >> n >> k;
    for(int i = 0; i < n; i++){
        T.a[i][i] = 1;
        for(int j = 0; j < n; j++){
            cin >> A.a[i][j];
        }
    }
    for(;k;k>>=1){
        if(k & 1) T = T * A;
        A = A * A;
        //T.show("T"); A.show("A");
    }
    ll ans = 0;
    for(int i = 0; i < n; i++) for(int j = 0; j < n; j++) add(ans, T.a[i][j]);
    //T.show();
    cout << ans << '\n';
}