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>

using namespace std;

typedef long long ll;
typedef long double ld;

const int mod = 1e9 + 7;
int n;

int add(int a, int b)
{
    return ((a % mod) + (b % mod)) % mod;
}

int mul(int a, int b)
{
    return ((ll)a * b) % mod;
}

vector<vector<int>> mult(vector<vector<int>> vec1, vector<vector<int>> vec2)
{
    vector<vector<int>> ret(n);
    for(int i = 0; i < n; i++)
    {
        ret[i].resize(n);
        for(int j = 0; j < n; j++)
        {
            for(int z = 0; z < n; z++){
                ret[i][j] = add(ret[i][j], mul(vec1[i][z], vec2[z][j]));
            }
        }
    }
    return ret;
}

vector<vector<int>> fastpow(vector<vector<int>> mat, ll power)
{
    if(!power)
    {
        vector<vector<int>> ret(n);
        for(int i = 0; i < n; i++)
        {
            ret[i].resize(n);
            ret[i][i] = 1;
        }
        return ret;
    }
    vector<vector<int>> ret = fastpow(mat, power / 2);
    ret = mult(ret, ret);
    if(power & 1)
        ret = mult(ret, mat);
    return ret;
}

int main()
{
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    ll k;
    cin >> n >> k;
    vector<vector<int>> mat(n);
    for(auto &x: mat){
        x.resize(n);
        for(auto &y: x)
            cin >> y;
    }
    vector<vector<int>> v = fastpow(mat, k);
    int ans = 0;
    for(int i = 0; i < n; i++)
        for(int j = 0; j < n; j++)
            ans = add(ans, v[i][j]);
    cout << ans;
}