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: Wrong Answer)

#include <bits/stdc++.h>
#include <algorithm>
#include <cmath>
#include <set>
#include <cstdio>
#include <vector>
#include <iostream>
#include <utility>
#include <queue>
#include <map>

#define fir first
#define sec second
#define sz(s) (s).size();
#define pb push_back
#define get(n) scanf("%d",&n);
#define gets(s) string s;cin >> (s);
#define prfi(n) printf("%d", &n);
#define prfd(n) printf("%lf", &n);
#define All(s) (s).begin(), (s).end()
#define rep(i,j,k) for(int (i)=(j);(i)<=(k);(i)++)
#define rep0(i,j) for(int (i)=0;(i)<(j);(i)++)
#define repdown(i, j, k) for(int (i)=(j);(i)>=(k);(i)--)
#define repdown0(i,j) for(int (i)=(j);i>=0;i--)
#define dump(x)  std::cout << #x << " = " << (x) << std::endl;
#define debug(x) cout << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;

using ll = long long;
using vi = std::vector<int> ;
using vvi = std::vector<vi> ;
using vll = std::vector<ll>;
using vvll = std::vector<vll>;
using vd = std::vector<double> ;
using vvd = std::vector<vd> ;
using qi = std::queue<int> ;
using vpii = std::vector<std::pair<int, int> >;
using namespace std;

const int Mod = (1e9) + 7;
const int max_n = 3 * (1e5) + 1;
const int max_m = 83 * (1e5) + 1;
const int INF = 10241024;
//const ll INFL = (1ll << 63) - 1;

//_____________________________________Templates_________________________________________//

template<class T1, class T2> inline void chmin(T1 &a, T2 b){if(a > b) a = b;}
template<class T1, class T2> inline void chmax(T1 &a, T2 b){if(a < b) a = b;}

//mainly use for dynamic prog
template<class T1, class T2>
void update(T1 &a, T2 b){
  a += b;
  if(a > Mod) a %= Mod;
}

inline void IN(void){
  return;
}

template <typename First, typename... Rest>
void IN(First& first, Rest&... rest){
  cin >> first;
  IN(rest...);
  return;
}

inline void OUT(void){
  cout << "\n";
  return;
}

template <typename First, typename... Rest>
void OUT(First first, Rest... rest){
  cout << first << " ";
  OUT(rest...);
  return;
}

//_____________________array calc____________________________________//

vvll mul(vvll &A, vvll &B){
  vvll C (A.size(), vll(B[0].size()));
  rep0(i,A.size()){
    rep0(j,B.size()){
      rep0(k,B[0].size()){
        C[i][j] += (C[i][j] + A[i][k] * B[k][j])%Mod;
      }
    }
  }
  return C;
}

vvll pow(vvll A, ll n){
  vvll B(A.size(), vll(A.size()));
  rep0(i,A.size()){
    B[i][i] = 1;
  }
  while(n > 0){
    if(n & 1)B = mul(B , A);
    A = mul(A, A);
    n >>= 1;
  }
  return B;
}
//_____________________Bynary Indexed Tree __________________________//
/*
const max_st = (1 << 15) - 1;
int bit[max_st];
int sum (int i){
  int s = 0;
  while(i > 0){
    s += bit[i];
    i -= i & -i;
  }
  return s;
}

void add(int i, int x){
  while(i <= n){
    bit[i] += x;
    i += i & -i;
  }
}
*/

//____________________RMQ____________________________//
/*
struct RMQ{
  #define s (1 << 20)
  ll dat[s];
  int n;

  void init(){
    memset(dat, -1e10, sizeof(dat));
  }

  void upd(int k, ll a){
    k += s/2 - 1;
    dat[k] = a;
    while(k > 0){
      k = (k -1) /2;
      dat[k] = max(dat[k*2+1], dat[k*2+2]);
    }
  }

  ll query(int a, int b, int k, int l, int r){
    if(r <= a || b <= l) return -1e8;
    if( a <= l && r <= b)return dat[k];
    else 
    {
      ll vl = query(a, b, k*2+1, l, (l+r)/2);
      ll vr = query(a, b, k*2+2, (l+r)/2, r);
      return max(vl, vr);
    }
  }

  void Debug(int a){
    a += s/2 - 1;
    rep(i,s/2 - 1,a){
      dump(dat[i]);
    }
  }
};
*/
//_____________________　following sorce code_________________________//
int n,m,k,x,y;
vvll dp(50, vll(50));
vvll v(50, vll(50));
ll ans = 0ll;

int main () {
  cin.tie(0);
  ios::sync_with_stdio(false);
  ll z;
  IN(n,z);
  
  rep0(i,n){
    rep0(j,n){
      int a;IN(a);
      v[i][j] = a;
    }
  }
  dp = pow(v, z);
  rep0(i,n){
    rep0(j,n){
      ans += dp[i][j];
      ans %= Mod;
    }
  }
  OUT(ans);
  return 0;
}