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 JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

import static java.lang.Integer.parseInt;

/**
 * One-Dimensional Cellular Automaton - Accepted
 */
public class Main {

	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String line;
		String[] words;

		while ((line = br.readLine()) != null && !line.isEmpty()) {

			StringTokenizer st = new StringTokenizer(line);

			int N, M, A, B, C, T;
			N = parseInt(st.nextToken());
			M = parseInt(st.nextToken());
			A = parseInt(st.nextToken());
			B = parseInt(st.nextToken());
			C = parseInt(st.nextToken());
			T = parseInt(st.nextToken());
			if ((N | M | A | B | C | T) == 0) break;

			//
			int[][] mat = new int[N][N];

			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;
			}

			int[][] aa = pow(mat, T, M);

			//
			st = new StringTokenizer(br.readLine());

			int[][] S = new int[N][1];

			for (int i = 0; i < N; i++) {
				S[i][0] = parseInt(st.nextToken());
			}

			S = mul(aa, S, M);

			String[] SS = new String[N];

			for (int i = 0; i < N; i++) {
				SS[i] = String.valueOf(S[i][0]);
			}

			System.out.println(String.join(" ", SS));
		}
	}

	static int[][] mul(int[][] a, int[][] b, int mod) {

		int[][] t = new int[a.length][b[0].length];

		for (int i = 0; i < a.length; i++) {
			for (int k = 0; k < b.length; k++) {
				for (int j = 0; j < b[0].length; j++) {
					t[i][j] = (t[i][j] + a[i][k] * b[k][j]) % mod;
				}
			}
		}

		return t;
	}

	static int[][] pow(int[][] a, int e, int mod) {

		if (e == 0) {
			int[][] t = new int[a.length][a[0].length];
			for (int i = 0; i < a.length; i++) {
				t[i][i] = 1;
			}
			return t;
		}

		if (e == 1) return a;

		int[][] t = {};

		t = pow(mul(a, a, mod), e / 2, mod);
		if (e % 2 == 1) t = mul(t, a, mod);

		return t;
	}
}