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.util.Scanner;

public class Main {
	static Scanner sc = new Scanner(System.in);
	static String g;
	static String[] a, b;
	static final int INF = 999;
	static int N, M, A, B, C, T;

	public static void main(String[] args) {
		while (true) {
			N = sc.nextInt();
			if (N == 0) break;
			M = sc.nextInt();
			A = sc.nextInt();
			B = sc.nextInt();
			C = sc.nextInt();
			T = sc.nextInt();
			int[] ans = solve();
			for (int i = 0; i < N; ++i) {
				if (i != 0) System.out.print(" ");
				System.out.print(ans[i]);
			}
			System.out.println();
		}
	}

	static int[] solve() {
		int[] v = new int[N];
		for (int i = 0; i < N; ++i) {
			v[i] = sc.nextInt();
		}
		int[][][] cnt = new int[30][N + 2][N + 2];
		for (int i = 1; i <= N; ++i) {
			cnt[0][i][i - 1] = C;
			cnt[0][i][i] = B;
			cnt[0][i][i + 1] = A;
		}
		for (int i = 1; (1 << i) <= T; ++i) {
			for (int j = 1; j <= N; ++j) {
				for (int k = 1; k <= N; ++k) {
					for (int l = 1; l <= N; ++l) {
						cnt[i][j][k] += cnt[i - 1][j][l] * cnt[i - 1][l][k];
					}
					cnt[i][j][k] %= M;
				}
			}
		}
		int[][] c = new int[2][N + 2];
		for (int i = 1; i <= N; ++i) {
			c[0][i] = v[i - 1];
		}
		int t = 0;
		for (int i = 0; (1 << i) <= T; ++i) {
			if ((T & (1 << i)) != 0) {
				for (int j = 1; j <= N; ++j) {
					c[1 - t][j] = 0;
				}
				for (int j = 1; j <= N; ++j) {
					for (int k = 1; k <= N; ++k) {
						c[1 - t][k] += c[t][j] * cnt[i][j][k];
					}
				}
				for (int j = 1; j <= N; ++j) {
					c[1 - t][j] %= M;
				}
				t = 1 - t;
			}
		}
		for (int i = 0; i < N; ++i) {
			v[i] = c[t][i + 1];
		}
		return v;
	}
}