Problem: 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

Solution (in CPP):

Solution:
#include<cstdio>
#include<vector>
using namespace std;
typedef long long int ll;
typedef vector<int> vec;
typedef vector<vec> mat;

int N, M;

mat mul(mat &A, mat &B){
	mat C(A.size(), vec(B[0].size()));
	for(int i = 0; i < A.size(); i++){
		for(int k = 0; k < B.size(); k++){
			for(int j = 0; j < B[0].size(); j++){
				C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % M;
			}
		}
	}
	return C;
}
mat pow(mat A, ll n){
	mat B(A.size(), vec(A.size()));
	for(int i = 0; i < A.size(); i++){
		B[i][i] = 1;
	}
	while(n > 0){
		if(n & 1) B = mul(B, A);
		A = mul(A, A);
		n >>= 1;
	}
	return B;
}

int main(){
	int A, B, C, T;
	for(;;){
		scanf("%d %d %d %d %d %d", &N, &M, &A, &B, &C, &T);
		if(N == 0 && M == 0 && A == 0 && B == 0 && C == 0 && T == 0) break;
		mat D(N, vec(N));
		for(int i = 0; i < N; i++){
			for(int j = 0; j < N; j++){
				if(i == j) D[i][j] = B;
				else if(i == j + 1) D[i][j] = C;
				else if(i == j - 1) D[i][j] = A;
				else D[i][j] = 0;
			}
		}
		mat x(1, vec(N));
		for(int i = 0; i < N; i++) scanf("%d", &x[0][i]);
		mat Dn = pow(D, T);
		mat res = mul(x, Dn);
		for(int i = 0; i < N; i++) printf("%d%c", res[0][i], (i == N - 1 ? '\n' : ' '));
	}
	return 0;
}