#include <bits/stdc++.h>
using namespace std;

#define for_(i,a,b) for(int i=(a);i<(b);++i)
template< typename T > using Vec = vector< T >;
template< typename T > T modAdd(T a, T b, T mod) { return (a+b)%mod; }
template< typename T > T modMul(T a, T b, T mod) { return (a*b)%mod; }

Vec< Vec< int > > mul(int N, int M, Vec< Vec< int > >& A, Vec< Vec< int > >& B) {
	Vec< Vec< int > > R(N, Vec< int >(N, 0));
	for_(i,0,N) for_(j,0,N) for_(k,0,N) R[i][j] = modAdd(R[i][j], modMul(A[i][k], B[k][j], M), M);
	return R;
}

Vec< Vec< int > > modPowMat(int N, int M, Vec< Vec< int > > A, int k) {
	Vec< Vec< int > > R(N, Vec< int >(N, 0));
	for_(i,0,N) R[i][i] = 1;
	while (k) {
		if (k & 1) R = mul(N, M, R, A);
		A = mul(N, M, A, A);
		k >>= 1;
	}
	return R;
}

int main() {
	int N, M, A, B, C, T;
	while (cin >> N >> M >> A >> B >> C >> T, N != 0) {
		Vec< int > S(N);
		for_(i,0,N) cin >> S[i];
		
		Vec< Vec< int > > mat(N, Vec< int >(N, 0));
		for_(i,0,N) {
			if (i - 1 >= 0) mat[i][i-1] = A;
			mat[i][i] = B;
			if (i + 1 < N) mat[i][i+1] = C;
		}
		
		mat = modPowMat(N, M, mat, T);
		
		for_(i,0,N) {
			int Si = 0;
			for_(j,0,N) Si = modAdd(Si, modMul(mat[i][j], S[j], M), M);
			cout << Si << (i < N-1 ? " " : "\n");
		}
	}
}
### Prompt

Develop a solution in cpp 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 <bits/stdc++.h>
using namespace std;

#define for_(i,a,b) for(int i=(a);i<(b);++i)
template< typename T > using Vec = vector< T >;
template< typename T > T modAdd(T a, T b, T mod) { return (a+b)%mod; }
template< typename T > T modMul(T a, T b, T mod) { return (a*b)%mod; }

Vec< Vec< int > > mul(int N, int M, Vec< Vec< int > >& A, Vec< Vec< int > >& B) {
	Vec< Vec< int > > R(N, Vec< int >(N, 0));
	for_(i,0,N) for_(j,0,N) for_(k,0,N) R[i][j] = modAdd(R[i][j], modMul(A[i][k], B[k][j], M), M);
	return R;
}

Vec< Vec< int > > modPowMat(int N, int M, Vec< Vec< int > > A, int k) {
	Vec< Vec< int > > R(N, Vec< int >(N, 0));
	for_(i,0,N) R[i][i] = 1;
	while (k) {
		if (k & 1) R = mul(N, M, R, A);
		A = mul(N, M, A, A);
		k >>= 1;
	}
	return R;
}

int main() {
	int N, M, A, B, C, T;
	while (cin >> N >> M >> A >> B >> C >> T, N != 0) {
		Vec< int > S(N);
		for_(i,0,N) cin >> S[i];
		
		Vec< Vec< int > > mat(N, Vec< int >(N, 0));
		for_(i,0,N) {
			if (i - 1 >= 0) mat[i][i-1] = A;
			mat[i][i] = B;
			if (i + 1 < N) mat[i][i+1] = C;
		}
		
		mat = modPowMat(N, M, mat, T);
		
		for_(i,0,N) {
			int Si = 0;
			for_(j,0,N) Si = modAdd(Si, modMul(mat[i][j], S[j], M), M);
			cout << Si << (i < N-1 ? " " : "\n");
		}
	}
}
```