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 pair<int,int> P;
typedef long long ll;
typedef long double ld;
const int inf=1e9+7;
const ll longinf=1LL<<60;
#define REP(i,m,n) for(int i=(int)(m) ; i < (int) (n) ; ++i )
#define rep(i,n) REP(i,0,n)
#define F first
#define S second
 
#define bit(n,k) ((n>>k)&1) //*n no k bit me 1 or 0*/

const int mx=100010;
const ll mod=1e9+7;

// matrix pow mod
void matpom(ll*mat,int size,ll n,int m){
  ll*temp=(ll*)malloc(size*size*sizeof(ll));
  ll*ans=(ll*)malloc(size*size*sizeof(ll));  
  rep(i,size)rep(j,size)ans[i*size+j] = i==j;
  while(n){
    if(n%2){
      rep(i,size*size)temp[i]=0;
      rep(i,size)rep(j,size)rep(k,size)temp[i*size+j]=(temp[i*size+j]+ans[i*size+k]*mat[k*size+j])%m;
      rep(i,size*size)ans[i]=temp[i];
    }
    rep(i,size*size)temp[i]=0;
    rep(i,size)rep(j,size)rep(k,size)temp[i*size+j]=(temp[i*size+j]+mat[i*size+k]*mat[k*size+j])%m;
    rep(i,size*size)mat[i]=temp[i];
    n/=2;
  }
  rep(i,size*size)mat[i]=ans[i];
  free(temp); free(ans);
}

ll a[50][50];         //in case of matrix pow mod //not vector
int main(){
  ll n,k;
  cin >> n >> k;
  rep(i,n)rep(j,n){ cin>>a[i][j]; }
  matpom((ll*)a,50,k,mod);
  ll ans = 0;
  rep(i,n)rep(j,n){ (ans+=a[i][j])%=mod; }
  cout << ans << endl;
  return 0;
}

/*
https://kyopro-friends.hatenablog.com/entry/2019/01/12/231035
dp[n][i][j]=(頂点iから頂点jへ行く長さnのパスの個数)
dp[n][i][j]=∑k dp[n−1][i][k]∗a[k][j]
となるわね。ところで、この積の形、よくみると行列の積と同じね！
dp[n][i][j]を(i,j)成分とする行列をDPnと書くことにすると、さっきの漸化式は
DPn=DPn−1A
になるわ。
DP0は明らかに単位行列だから、DPn=Anね。
求める答えは∑i,jdp[K][i][j]だから、AKを計算すれば求められるわよ。
行列累乗で解く問題は、次の２つのポイントを抑えておくといいわ。どっちも決め手になるほどじゃないけど……
・ポイント１：制約をみる
N≤50くらい、Kは109とか1018とかの大きな値、みたいな制約の問題は、サイズN×Nの行列をK乗するかも。
・ポイント２：繰り返し
今回の状態遷移の式を見ると、回数nが遷移に影響しないわね。
こういうふうに同じ状態遷移を何度も繰り返すものは、行列累乗かも。
*/