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)

// 解説みてかいた
#define DEBUG 0
#include <algorithm>
#include <cmath>
#include <cstring>
#include <deque>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <tuple>
#include <utility>
#include <vector>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using vll = vector<ll>;
using vvll = vector<vll>;
using pll = pair<ll, ll>;
using tll = tuple<ll, ll, ll>;
#define all(v) (v).begin(), (v).end()
#define for1(i, n) for (ll i = 0; i < (n); i++)
#define for2(i, m, n) for (ll i = (m); i < (n); i++)
#define for3(i, m, n, d) for (ll i = (m); i < (n); i += (d))
#define rfor2(i, m, n) for (ll i = (m); i > (n); i--)
#define rfor3(i, m, n, d) for (ll i = (m); i > (n); i += (d))
#define PI 3.1415926535897932384626433832795028841971693993751
#define INF 1111111111111111111LL
#define print(...) print_1(__VA_ARGS__)
#define in(...) in_1(__VA_ARGS__)
#if DEBUG
#define dump(...) dump_1(#__VA_ARGS__, __VA_ARGS__)
#define dumpa(...) dumpa_1(#__VA_ARGS__, __VA_ARGS__)
#else
#define dump(...)
#define dumpa(...)
#endif
template <typename Head>
void dump_1(const char* str, Head&& h)
{
    cerr << str << ": " << h << '\n';
}
template <typename Head, typename... Tail>
void dump_1(const char* str, Head&& h, Tail&&... t)
{
    while (*str != ',') {
        cerr << *str++;
    }
    cerr << ": " << h << ' ';
    dump_1(str + 1, t...);
}
template <typename T>
void dumpa_1(const char* str, const T v[], const ll size)
{
    while (*str != ',') {
        cerr << *str++;
    }
    cerr << ": ";
    for1 (i, size) {
        if (i != 0) {
            cerr << ' ';
        }
        cerr << v[i];
    }
    cerr << '\n';
}
template <typename T1, typename T2>
ostream& operator<<(ostream& os, const pair<T1, T2>& v)
{
    os << '(' << v.first << ", " << v.second << ')';
    return os;
}
template <typename T1, typename T2, typename T3>
ostream& operator<<(ostream& os, const tuple<T1, T2, T3>& v)
{
    os << '(' << get<0>(v) << ", " << get<1>(v) << ", " << get<2>(v) << ')';
    return os;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v)
{
    for (auto it = v.begin(); it != v.end(); it++) {
        if (it != v.begin()) {
            os << ' ';
        }
        os << *it;
    }
    return os;
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& v)
{
    for (auto it = v.begin(); it != v.end(); it++) {
        if (it != v.begin()) {
            os << ' ';
        }
        os << *it;
    }
    return os;
}
template <typename T>
ostream& operator<<(ostream& os, const multiset<T>& v)
{
    for (auto it = v.begin(); it != v.end(); it++) {
        if (it != v.begin()) {
            os << ' ';
        }
        os << *it;
    }
    return os;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& os, const map<T1, T2>& v)
{
    os << '{';
    for (auto it = v.begin(); it != v.end(); it++) {
        if (it != v.begin()) {
            os << ", ";
        }
        os << it->first << ':' << it->second;
    }
    os << '}';
    return os;
}
void Yes(void) { cout << "Yes\n"; }
void No(void) { cout << "No\n"; }
void YES(void) { cout << "YES\n"; }
void NO(void) { cout << "NO\n"; }
template <typename T>
bool chmax(T& a, const T& b)
{
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
template <typename T>
bool chmin(T& a, const T& b)
{
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template <typename T>
void vin(vector<T>& v, ll len)
{
    for1 (i, len) {
        cin >> v[i];
    }
}
template <typename Head>
void in_1(Head& h)
{
    cin >> h;
}
template <typename Head, typename... Tail>
void in_1(Head& h, Tail&... t)
{
    cin >> h;
    in_1(t...);
}
template <typename Head>
void print_1(Head&& h)
{
    cout << h << '\n';
}
template <typename Head, typename... Tail>
void print_1(Head&& h, Tail&&... t)
{
    cout << h << ' ';
    print_1(t...);
}
//---------------------------------------------------------
//---------------------------------------------------------
const ll mod = 1e9 + 7;
struct mint {
    ll x;  // typedef long long ll;
    mint(ll x = 0) : x((x % mod + mod) % mod) {}
    mint& operator+=(const mint a)
    {
        if ((x += a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator-=(const mint a)
    {
        if ((x += mod - a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator*=(const mint a)
    {
        (x *= a.x) %= mod;
        return *this;
    }
    mint operator+(const mint a) const
    {
        mint res(*this);
        return res += a;
    }
    mint operator-(const mint a) const
    {
        mint res(*this);
        return res -= a;
    }
    mint operator*(const mint a) const
    {
        mint res(*this);
        return res *= a;
    }
    mint pow(ll t) const
    {
        if (!t) return 1;
        mint a = pow(t >> 1);
        a *= a;
        if (t & 1) a *= *this;
        return a;
    }

    // for prime mod
    mint inv() const { return pow(mod - 2); }
    mint& operator/=(const mint a) { return (*this) *= a.inv(); }
    mint operator/(const mint a) const
    {
        mint res(*this);
        return res /= a;
    }
    friend ostream& operator<<(ostream& os, const mint& v)
    {
        os << v.x;
        return os;
    }
};
//---------------------------------------------------------
vector<vector<mint>> mul_matrix(vector<vector<mint>> A, vector<vector<mint>>& B,
                                ll N)
{
    vector<vector<mint>> ret(N, vector<mint>(N));
    for1 (i, N) {
        for1 (j, N) {
            for1 (k, N) {
                ret[i][j] += A[i][k] * B[k][j];
            }
        }
    }
    return ret;
}
void solve()
{
    ll N, K;
    in(N, K);
    vector<vector<mint>> A(N, vector<mint>(N));
    for1 (i, N) {
        for1 (j, N) {
            ll x;
            in(x);
            A[i][j] = x;
        }
    }

    vector<vector<vector<mint>>> A2(64);
    A2[0] = A;
    for2 (i, 1, 64) {
        A2[i] = mul_matrix(A2[i - 1], A2[i - 1], N);
    }

    vector<vector<mint>> A3 = A2[0];
    K--;
    for1 (i, 64) {
        if (((ull)K >> i) & 1) {
            dump(i);
            A3 = mul_matrix(A3, A2[i], N);
        }
    }
    mint ans = 0;
    for1 (i, N) {
        for1 (j, N) {
            ans += A3[i][j];
        }
    }
    for1 (i, N) {
        dump(A3[i]);
    }
    print(ans);
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << fixed << setprecision(16);
    cerr << fixed << setprecision(16);
    solve();
}