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 <iostream>
#include <iomanip>
#include <sstream>
#include <vector>
#include <string>
#include <set>
#include <unordered_set>
#include <map>
#include <unordered_map>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <functional>
#include <iterator>
#include <limits>
#include <numeric>
#include <utility>
#include <type_traits>
#include <cmath>
#include <cassert>
#include <cstdio>

using namespace std;
using namespace placeholders;

using LL = long long;
using ULL = unsigned long long;
using VI = vector< int >;
using VVI = vector< vector< int > >;
using VS = vector< string >;
using ISS = istringstream;
using OSS = ostringstream;
using PII = pair< int, int >;
using VPII = vector< pair< int, int > >;
template < typename T = int > using VT = vector< T >;
template < typename T = int > using VVT = vector< vector< T > >;
template < typename T = int > using LIM = numeric_limits< T >;

template < typename T > inline istream& operator>>( istream &s, vector< T > &v ){ for ( T &t : v ) { s >> t; } return s; }
template < typename T > inline ostream& operator<<( ostream &s, const vector< T > &v ){ for ( int i = 0; i < int( v.size() ); ++i ){ s << ( " " + !i ) << v[i]; } return s; }

void in_impl(){};
template < typename T, typename... TS > void in_impl( T &head, TS &... tail ){ cin >> head; in_impl( tail ... ); }
#define IN( T, ... ) T __VA_ARGS__; in_impl( __VA_ARGS__ );

template < typename T > struct getv_fmt;
template <> struct getv_fmt<       int >{ static constexpr const char *fmt = "%d"; };
template <> struct getv_fmt< long long >{ static constexpr const char *fmt = "%lld"; };
template < typename T > void getv( std::vector< T > &v ){ for_each( begin( v ), end( v ), []( T &a ){ scanf( getv_fmt< T >::fmt, &a ); } ); };

template < typename T > inline T fromString( const string &s ) { T res; istringstream iss( s ); iss >> res; return res; }
template < typename T > inline string toString( const T &a ) { ostringstream oss; oss << a; return oss.str(); }

#define NUMBERED( name, number ) NUMBERED2( name, number )
#define NUMBERED2( name, number ) name ## _ ## number
#define REP1( n ) REP2( NUMBERED( REP_COUNTER, __LINE__ ), n )
#define REP2( i, n ) REP3( i, 0, n )
#define REP3( i, m, n ) for ( int i = ( int )( m ); i < ( int )( n ); ++i )
#define GET_REP( a, b, c, F, ... ) F
#define REP( ... ) GET_REP( __VA_ARGS__, REP3, REP2, REP1 )( __VA_ARGS__ )
#define FOR( e, c ) for ( auto &&e : c )
#define ALL( c ) begin( c ), end( c )
#define AALL( a ) ( remove_all_extents< decltype( a ) >::type * )a, ( remove_all_extents< decltype( a ) >::type * )a + sizeof( a ) / sizeof( remove_all_extents< decltype( a ) >::type )
#define DRANGE( c, p ) begin( c ), begin( c ) + ( p ), end( c )

#define SZ( v ) ( (int)( v ).size() )
#define EXIST( c, e ) ( ( c ).find( e ) != ( c ).end() )

template < typename T > inline bool chmin( T &a, const T &b ){ if ( b < a ) { a = b; return true; } return false; }
template < typename T > inline bool chmax( T &a, const T &b ){ if ( a < b ) { a = b; return true; } return false; }

#define PB push_back
#define EM emplace
#define EB emplace_back
#define BI back_inserter

#define MP make_pair
#define fst first
#define snd second

#define DUMP( x ) cerr << #x << " = " << ( x ) << endl

constexpr int MOD = 1'000'000'007;

template < typename Number > using Vector = vector< Number >;
template < typename Number > using Matrix = vector< vector< Number > >;

// 行列積のためのヘルパ関数
double modulo( const double a, const long long mod )
{
	return a;
}
int modulo( const LL a, const long long mod )
{
	return a % mod;
}

// 単位行列
// O( N )
template < typename Number >
Matrix< Number > identify_matrix( const int N )
{
	Matrix< Number > res( N, Vector< Number >( N ) );
	for ( int i = 0; i < N; ++i )
	{
		res[i][i] = 1;
	}
	return res;
}

// 行列同士の積
// O( N^3 )
template < typename Number >
Matrix< Number > matrix_multiply( const Matrix< Number > &a, const Matrix< Number > &b, unsigned int mod = 0 )
{
	assert( a[0].size() == b.size() );

	Matrix< Number > res( a.size(), Vector< Number >( b[0].size() ) );
	for ( int i = 0; i < static_cast<int>( a.size() ); ++i )
	{
		for ( int j = 0; j < static_cast<int>( b[0].size() ); ++j )
		{
			for ( int k = 0; k < static_cast<int>( b.size() ); ++k )
			{
				res[i][j] += a[i][k] * b[k][j];
				res[i][j] = modulo( res[i][j], mod );
			}
		}
	}
	return res;
}
template < typename Number >
inline const Matrix< Number > operator*( const Matrix< Number > &a, const Matrix< Number > &b )
{
	return matrix_multiply( a, b );
}

// 行列累乗
// OO( N^3 log n )
template < typename Number >
Matrix< Number > power( Matrix< Number > a, long long n, long long mod = numeric_limits< long long >::max() )
{
	Matrix< Number > res = identify_matrix< Number >( a.size() );

	while ( n )
	{
		if ( n & 1 )
		{
			res = matrix_multiply( res, a, mod 	);
		}
		a = matrix_multiply( a, a, mod );
		n >>= 1;
	}

	return res;
}

int main()
{
	cin.tie( 0 );
	ios::sync_with_stdio( false );
	cout << setprecision( 12 ) << fixed;

	IN( int, N );
	IN( LL, K );
	Matrix< LL > G( N, Vector< LL >( N ) );
	cin >> G;

	G = power( G, K, MOD );

	int res = 0;
	REP( i, N )
	{
		REP( j, N )
		{
			( res += G[i][j] ) %= MOD;
		}
	}

	cout << res << endl;

	return 0;
}