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 ll long long
#define ld long double
#define mk make_pair
#define fi first
#define se second
#define vll vector<ll>
#define pii pair<ll,ll>
#define vvll vector< vector<ll> >
#define pb push_back
#define sz(v) (v).size()
#define inf 1e16
#define md 1000000007
#define all(v) (v).begin(),(v).end()
#define rep(i,a,b) for(ll i=a;i<b;++i)
#define tel(a){cout<<a<<"\n";}
#define tell(a,b){cout<<a<<" | "<<b<<"\n";}
#define telll(a,b,c){cout<<a<<" | "<<b<<" | "<<c<<"\n";}
#define teln(v,n){cout<<"v- ";rep(i,0,n)cout<<v[i]<<" ";cout<<"\n";}
#define IOS ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace std;
#define M 1000010
const int N=51;

void print(ll a[][N],ll n){rep(i,0,n){rep(j,0,n)cout<<a[i][j]<<" ";cout<<"\n";}}
void id(ll a[][N],ll n){rep(i,0,n)rep(j,0,n){a[i][j]=(i==j);}}
void mul(ll a[][N],ll b[][N],ll c[][N],ll n){rep(i,0,n)rep(j,0,n){c[i][j]=0;rep(k,0,n)c[i][j]=(c[i][j]+a[i][k]*b[k][j])%md;}}

void exp(ll a[][N],ll k,ll n){
	ll temp[N][N],ans[N][N];id(ans,n);
	while(k){
		if(k&1){
			mul(ans,a,temp,n);
			rep(i,0,n)rep(j,0,n)ans[i][j]=temp[i][j];
		}
		mul(a,a,temp,n);
		rep(i,0,n)rep(j,0,n)a[i][j]=temp[i][j];
		k/=2;
	}
	rep(i,0,n)rep(j,0,n)a[i][j]=ans[i][j];
}

int main(){
	IOS;
	ll n,k;cin>>n>>k;
	ll a[N][N];
	rep(i,0,n)rep(j,0,n)cin>>a[i][j];exp(a,k,n);
	ll ans=0;rep(i,0,n)rep(j,0,n)ans=(ans+a[i][j])%md;
	cout<<ans;
	
	return 0;
}