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)

/*
___mennat khoday ra azz o va jall ke taatash mojebe ghorbat ast o be shokr andarash mazide nemat.
___har nafasi ke foro miravad momedde hayatast o chon bar miayad mofarrahe zat.
___pas dar har nafasi 2 nemat mojod ast o bar har nematy shokry vajeb.

___az dast o zabane ke barayad ?
___kaz ohdeye shokrash be dar ayad !


___man be to dami gharar natvanam kard
___va ehsane to ra shomar natvanam kard
####################
___gar bar tane man zaban shavad har moyee
___yek shokre to az hezar natvanam kard

*/

#include <bits/stdc++.h>

#define pb push_back
#define SALI main
#define left(x) ((x)*2)
#define right(x) (((x)*2)+1)
#define fuck(x) cout << #x << " : " << x << endl;
#define findLowVec(v,x) (lower_bound(v.begin(), v.end(), x) - v.begin())
#define findUpVec(v,x) (upper_bound(v.begin(), v.end(), x) - v.begin())
#define findLowArr(a,n,x) (lower_bound(a, a+n, x) - a)
#define findUpArr(a,n,x) (upper_bound(a, a+n, x) - a)
#define int long long

using namespace std;

typedef long long ll;
typedef pair<int,int> pii;
typedef long double ld;

template<class A, class B> ostream& operator <<(ostream& out, const pair<A, B> &p)
{return out << "(" << p.first << ", " << p.second << ")";}

template<class A> ostream& operator <<(ostream& out, const vector<A> &v)
{out << "[";for(int i = 0; i < v.size(); i++) {if(i) out << ", ";out << v[i];}return out << "]";}

template<class A,class B> ostream& operator <<(ostream& out, const set<A,B> &s)
{out << "[";for(auto i = s.begin(); i != s.end(); i++) {out << ", ";out << *i;}return out << "]";}

template<class A> ostream& operator <<(ostream& out, const set<A> &s)
{out << "[";for(auto i = s.begin(); i != s.end(); i++) {out << ", ";out << *i;}return out << "]";}

const int N = 52;
const int mod = 1e9 + 7;

struct matrix
{

    int a[N][N];

    matrix(int idd = 0){
        for(int i = 0;i<N;i++){
            for(int j = 0;j<N;j++){
                a[i][j] = (idd) ? (i == j) : (0);
            }
        }
    }

    matrix operator*(const matrix& other)const{
        matrix ans;

        for(int i = 0; i < N; i++){
            for(int j = 0; j < N; j++){
                for(int k = 0; k < N; k++){
                    ans.a[i][j] += a[i][k] * other.a[k][j];
                    ans.a[i][j] %= mod;
                }
            }
        }
        return ans;
    }

};

matrix power(matrix base, int k)
{
    matrix ans(1);

    for(int i = 0;(1LL << i) <= k;i++){
        if((1LL << i) & k) ans = ans*base;
        base = base*base;
    }

    return ans;
}

int32_t SALI()
{
    int n,k;
    cin >> n >> k;
    matrix a;
    for(int i = 0; i < n; i++) for(int j = 0; j < n; j++) cin >> a.a[i][j];
    matrix ans = power(a,k);
    int tot = 0;
    for(int i = 0; i < n; i++){
        for(int j = 0; j < n; j++){
            tot += ans.a[i][j];
            tot %= mod;
        }
    }

    cout << tot;


}




/**< WRITEN BY ALI ADELKHAH */