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 <stdio.h>
#include <memory.h>
#include <vector>

using namespace std;

typedef vector<long long> Array;
typedef vector<Array> Matrix;

//FILE *in = freopen("input.txt", "r", stdin);
//FILE *out = freopen("output.txt", "w", stdout);

int N, M, A, B, C, T;
long long S[100];
long long res[100];

Matrix mat;

void input() {
	memset(S, 0, sizeof(S));
	memset(res, 0, sizeof(res));
	mat.clear();
	for(int i = 0; i < N; i++) {
		scanf("%d", &S[i]);
		mat.push_back(Array(N));
		mat[i][i] = 1;
	}
}

Matrix mul(const Matrix &a, const Matrix &b) {
	Matrix res(a.size(), Array(b[0].size()));
	for(int i = 0; i < a.size(); i++) {
		for(int j = 0; j < b[0].size(); j++) {
			for(int k = 0; k < a[0].size(); k++) {
				res[i][j] = (res[i][j] + a[i][k]*b[k][j])%M;
			}
		}
	}
	return res;
}

Matrix matmul(int n) {
	if(n == 1)
		return mat;
	Matrix res = matmul(n/2);
	res = mul(res, res);
	if(n&1)
		res = mul(res, mat);
	return res;
}

void process() {
	for(int i = 0; i < N; i++) {
		if(i-1 >= 0) 
			mat[i][i-1] = A;
		mat[i][i] = B;
		if(i+1 < N)
			mat[i][i+1] = C;
	}

	mat = matmul(T);
}

void output() {
	for(int i = 0; i < N; i++) {
		for(int j = 0; j < N; j++) {
			res[i] = (res[i] + mat[i][j]*S[j])%M;
		}
	}

	for(int i = 0; i < N - 1; i++) {
		printf("%Ld ", res[i]);
	}
	printf("%Ld\n", res[N-1]);
}

int main() {
	while(true){
		scanf("%d%d%d%d%d%d", &N, &M, &A, &B, &C, &T);
		if(N == 0) break;
		input();
		if(T)
			process();
		output();
	}
	return 0;
}