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)

#include <bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;   
#define ordered_set tree<ll, null_type,less_equal<ll>, rb_tree_tag,tree_order_statistics_node_update> 
#define PI 3.14159265
#define ll long long
#define ld long double
#define vi vector<int>
#define pb push_back
#define vll vector<ll>
#define vvi vector < vi >
#define pii pair<int,int>
#define pll pair<long long, long long>
#define mod 1000000007
#define inf 300000000000000001
#define infx 9223372036854775806
#define all(c) c.begin(),c.end()
#define mk(x,yy) make_pair(x,y)
#define mem(a,val) memset(a,val,sizeof(a))
#define eb emplace_back
#define ff first
#define ss second
#define re return
// #define endl "\n"
#define max2(x,y)           (x>y)?x:y
#define min2(x,y)           (x<y)?x:y
#define mid(s,e)               (s+(e-s)/2)
#define mini                   INT_MIN
#define maxo                   INT_MAX
#define rep(i,a,b) for(int i=a;i<(int)(b);++i)
#define read(a,n) for(int i=0;i<n;i++)cin>>a[i]
#define show(a,n) for(int i=0;i<n;i++)cout<<a[i]<<' ';cout<<endl
#define ene cout<<"\n";

ll gcd(ll a, ll b) 
{ 
    if (b == 0) 
        return a; 
    return gcd(b, a % b);  
      
}
ll power(ll x,  ll y, ll p) 
{ 
    ll res = 1;      // Initialize result 
  
    x = x % p;  // Update x if it is more than or  
                // equal to p 
  
    while (y > 0) 
    { 
        // If y is odd, multiply x with result 
        if (y & 1) 
            res = (res*x) % p; 
  
        // y must be even now 
        y = y>>1; // y = y/2 
        x = (x*x) % p;   
    } 
    return res; 
} 
 
bool compr(std::vector<ll> a, std::vector<ll> b)
{
    if(a[0]==b[0])
        return a[1]<b[1];
    else
        return a[0]<b[0];
}

bool comp(std::vector<ll> a, std::vector<ll> b)
{
    if(a[1]==b[1]&&a[2]==b[2])
        return a[0]<b[0];
    else if(a[1]==b[1])
        return a[2]>b[2];
    else
        return a[1]<b[1];
}
void add_self(int&a, int b) {
    a += b;
    if(a >= mod) {
        a -= mod;
    }
}
int mul(int a, int b) {
    return (ll) a * b % mod;
}

int n;

struct M {
    vector<vector<int>> t;
    M() {
        t.resize(n, vector<int>(n));
    }
    M operator* (const M& b) const {
        M c = M();
        for(int i = 0; i < n; ++i) {
            for(int j = 0; j < n; ++j) {
                for(int k = 0; k < n; ++k) {
                    add_self(c.t[i][k], mul(t[i][j], b.t[j][k]));
                }
            }
        }
        return c;
    }
};
int main()
{
    #ifdef SHIVANSH 
     freopen("input.txt","r",stdin);
     freopen("output.txt","w",stdout);
    #endif
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll k;
    scanf("%d%lld", &n, &k);
    vector<vector<int>> can(n, vector<int>(n));
    for(int i = 0; i < n; ++i) {
        for(int j = 0; j < n; ++j) {
            scanf("%d", &can[i][j]);
        }
    }
    M answer = M();
    for(int i = 0; i < n; ++i) {
        answer.t[i][i] = 1;
    }
    M m = M();
    m.t = can;
    while(k) {
        if(k % 2) {
            answer = answer * m;
        }
        m = m * m;
        k /= 2;
    }
    int total = 0;
    for(int i = 0; i < n; ++i) {
        for(int j = 0; j < n; ++j) {
            add_self(total, answer.t[i][j]);
        }
    }
    printf("%d\n", total);
    return 0;
    
}