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
#include <cstdio>
#include <vector>
using namespace std;

typedef vector<vector<int> > vvint;

void mul(vvint &x, const vvint &y, int m){
	int n = x.size();
	vvint z(n, vector<int>(n));
	for(int i = 0; i < n; ++i)
		for(int j = 0; j < n; ++j)
			for(int k = 0; k < n; ++k)
				z[i][j] = (z[i][j] + (long long)x[i][k] * y[k][j]) % m;
	x.swap(z);
}

int main(){
	int n, m, a, b, c, t;
	while(scanf("%d%d%d%d%d%d", &n, &m, &a, &b, &c, &t), n){
		vvint x(n, vector<int>(n)), y = x;
		for(int i = 0; i < n; ++i){
			if(i != 0) x[i][i-1] = a;
			x[i][i] = b;
			if(i != n - 1) x[i][i+1] = c;

			y[i][i] = 1;
		}

		while(t){
			if(t & 1) mul(y, x, m);
			mul(x, x, m);
			t >>= 1;
		}

		vector<int> s(n);
		for(int i = 0; i < n; ++i){
			scanf("%d", &s[i]);
		}

		for(int i = 0; i < n; ++i){
			int p = 0;
			for(int j = 0; j < n; ++j){
				p = (p + (long long)y[i][j] * s[j]) % m;
			}
			printf("%d%c", p, i == n - 1 ? '\n' : ' ');
		}
	}
}