
#include <iostream>
#include <vector>

using namespace std;
typedef vector<vector<int>> matrix;
typedef vector<int> vec;

matrix matprod( matrix &A, matrix &B, int mod = 1);
matrix matpow( matrix src, int p, int M );
int solve();

int main(void) {
	
	while(solve()){}

	return 0;
}

matrix matprod( matrix &A, matrix &B, int mod ) {

	matrix ret( B.size(), vec( A[0].size(), 0 ) );
	for( int i = 0; i < ret.size(); i++ ) {
		for( int j = 0; j < ret[i].size(); j++ ) {
			for( int k = 0; k < A[i].size(); k++ ) {
				ret[i][j] += A[i][k]*B[k][j];
				ret[i][j] %= mod;
			}
		}
	}

	return ret;

}
matrix matpow( matrix src, int p, int M) {

	//???????´????
	if( p == 0 ) {
		matrix ret(src);
		for( int i = 0; i < ret.size(); i++ ) {
			for( int j = 0; j < ret[i].size(); j++ ) {
				if( i == j ) {
					ret[i][j] = 1;
				} else {
					ret[i][j] = 0;
				}
			}
		}
		return ret;
	}
	if( p == 1 ) { return src; }

	if( p & 1 ) {

		matrix n;
		matrix m(matpow(src,(p-1)/2,M));

		n = matprod(m,m,M);
		return matprod(src,n,M);

	} else {

		matrix m(matpow(src,p/2,M));
		return matprod(m,m,M);

	}

}

int solve() {

	int N,M,A,B,C,T;

	cin >> N >> M >> A >> B >> C >> T;

	if( N == 0 ){ return 0; }

	matrix mat( N, vec(N, 0) );
	matrix ans( N, vec(N, 0) );
	matrix mat2(N, vec(N, 0) );

	vector<int> S(N,0);

	int t[3] = { A, B, C };

	for( int i = 0; i < N; i++ )
		{ cin >> S[i]; }

	for( int i = 0; i < N; i++ ) {
		for( int k = -1; k < 2; k++ ) {
			if( i + k < 0 || i + k >= N ) { continue; }
			mat[i][i+k] = t[k+1];
		}
	}
	for( int i = 0; i < N; i++ ) {
		for( int j = 0; j < N; j++ ) {
			mat2[i][j] = S[(i+j)%N];
		}
	}

	mat = matpow( mat, T, M );
	ans = matprod( mat, mat2, M );
	
	for( int i = 0; i < N; i++ ) {
		if( i > 0 ) { cout << " "; }
		cout << ans[i][0];
	}
	
	cout << endl;
	
	return 1;

}
### Prompt

Please provide a cpp coded solution to the problem described below:

There is a one-dimensional cellular automaton consisting of N cells. Cells are numbered from 0 to N − 1.

Each cell has a state represented as a non-negative integer less than M. The states of cells evolve through discrete time steps. We denote the state of the i-th cell at time t as S(i, t). The state at time t + 1 is defined by the equation

S(i, t + 1) = (A × S(i − 1, t) + B × S(i, t) + C × S(i + 1, t)) mod M,         (1)

where A, B and C are non-negative integer constants. For i < 0 or N ≤ i, we define S(i, t) = 0.

Given an automaton definition and initial states of cells, your mission is to write a program that computes the states of the cells at a specified time T.



Input

The input is a sequence of datasets. Each dataset is formatted as follows.

N M A B C T
S(0, 0) S(1, 0) ... S(N − 1, 0)


The first line of a dataset consists of six integers, namely N, M, A, B, C and T. N is the number of cells. M is the modulus in the equation (1). A, B and C are coefficients in the equation (1). Finally, T is the time for which you should compute the states.

You may assume that 0 < N ≤ 50, 0 < M ≤ 1000, 0 ≤ A, B, C < M and 0 ≤ T ≤ 109.

The second line consists of N integers, each of which is non-negative and less than M. They represent the states of the cells at time zero.

A line containing six zeros indicates the end of the input.

Output

For each dataset, output a line that contains the states of the cells at time T. The format of the output is as follows.

S(0, T) S(1, T) ... S(N − 1, T)

Each state must be represented as an integer and the integers must be separated by a space.

Example

Input

5 4 1 3 2 0
0 1 2 0 1
5 7 1 3 2 1
0 1 2 0 1
5 13 1 3 2 11
0 1 2 0 1
5 5 2 0 1 100
0 1 2 0 1
6 6 0 2 3 1000
0 1 2 0 1 4
20 1000 0 2 3 1000000000
0 1 2 0 1 0 1 2 0 1 0 1 2 0 1 0 1 2 0 1
30 2 1 0 1 1000000000
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
30 2 1 1 1 1000000000
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
30 5 2 3 1 1000000000
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0


Output

0 1 2 0 1
2 0 0 4 3
2 12 10 9 11
3 0 4 2 1
0 4 2 0 4 4
0 376 752 0 376 0 376 752 0 376 0 376 752 0 376 0 376 752 0 376
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0
1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 3 2 2 2 3 3 1 4 3 1 2 3 0 4 3 3 0 4 2 2 2 2 1 1 2 1 3 0

### Response

```cpp

#include <iostream>
#include <vector>

using namespace std;
typedef vector<vector<int>> matrix;
typedef vector<int> vec;

matrix matprod( matrix &A, matrix &B, int mod = 1);
matrix matpow( matrix src, int p, int M );
int solve();

int main(void) {
	
	while(solve()){}

	return 0;
}

matrix matprod( matrix &A, matrix &B, int mod ) {

	matrix ret( B.size(), vec( A[0].size(), 0 ) );
	for( int i = 0; i < ret.size(); i++ ) {
		for( int j = 0; j < ret[i].size(); j++ ) {
			for( int k = 0; k < A[i].size(); k++ ) {
				ret[i][j] += A[i][k]*B[k][j];
				ret[i][j] %= mod;
			}
		}
	}

	return ret;

}
matrix matpow( matrix src, int p, int M) {

	//???????´????
	if( p == 0 ) {
		matrix ret(src);
		for( int i = 0; i < ret.size(); i++ ) {
			for( int j = 0; j < ret[i].size(); j++ ) {
				if( i == j ) {
					ret[i][j] = 1;
				} else {
					ret[i][j] = 0;
				}
			}
		}
		return ret;
	}
	if( p == 1 ) { return src; }

	if( p & 1 ) {

		matrix n;
		matrix m(matpow(src,(p-1)/2,M));

		n = matprod(m,m,M);
		return matprod(src,n,M);

	} else {

		matrix m(matpow(src,p/2,M));
		return matprod(m,m,M);

	}

}

int solve() {

	int N,M,A,B,C,T;

	cin >> N >> M >> A >> B >> C >> T;

	if( N == 0 ){ return 0; }

	matrix mat( N, vec(N, 0) );
	matrix ans( N, vec(N, 0) );
	matrix mat2(N, vec(N, 0) );

	vector<int> S(N,0);

	int t[3] = { A, B, C };

	for( int i = 0; i < N; i++ )
		{ cin >> S[i]; }

	for( int i = 0; i < N; i++ ) {
		for( int k = -1; k < 2; k++ ) {
			if( i + k < 0 || i + k >= N ) { continue; }
			mat[i][i+k] = t[k+1];
		}
	}
	for( int i = 0; i < N; i++ ) {
		for( int j = 0; j < N; j++ ) {
			mat2[i][j] = S[(i+j)%N];
		}
	}

	mat = matpow( mat, T, M );
	ans = matprod( mat, mat2, M );
	
	for( int i = 0; i < N; i++ ) {
		if( i > 0 ) { cout << " "; }
		cout << ans[i][0];
	}
	
	cout << endl;
	
	return 1;

}
```