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.*;

public class Main{
	int m;
	private void doit(){
		Scanner sc = new Scanner(System.in);
		while(true){
			int n = sc.nextInt();
			m = sc.nextInt();
			int a = sc.nextInt();
			int b = sc.nextInt();
			int c = sc.nextInt();
			int t = sc.nextInt();
			if((n|m|a|b|c|t) == 0){
				break;
			}
			int [][] s = new int[n][1];
			for(int i = 0; i < n; i++){
				s[i][0] = sc.nextInt();
			}
			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 [][] matrixpowmat = matrixPow(mat, t);
			int [][] ans = mul(matrixpowmat, s);
			for(int i = 0; i < n-1; i++){
				System.out.print(ans[i][0] + " ");
			}
			System.out.println(ans[n-1][0]);
		}
	}

	private int[][] matrixPow(int[][] a, int n) {
		int [][] b = new int [a.length][a[0].length];
		for(int i = 0; i < a.length; i++){
			b[i][i] = 1;
		}
		while(n > 0){
			if((n & 1) != 0){
				b = mul(b,a);
			}
			a = mul(a,a);
			n >>= 1;
		}
		return b;
	}

	private int[][] mul(int[][] a, int[][] b) {
		int [][] c = 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++){
					c[i][j] = (c[i][j] + a[i][k] * b[k][j]) % m;
				}
			}
		}
		return c;
	}

	private void debug(Object... o) {
		System.out.println("debug = " + Arrays.deepToString(o));
	}

	public static void main(String[] args) {
		new Main().doit();
	}
}