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.*;
import java.io.*;
import static java.util.Arrays.*;
import static java.util.Collections.*;
import static java.lang.Math.*;

public class Main {
	int INF = 1<<28;
	double EPS = 1e-9;
	long MOD = (long)1e9 + 7;
	int n, m;
	void run() {
		Scanner sc = new Scanner(System.in);
		for(;;) {
			n = sc.nextInt(); m = sc.nextInt();
			int a = sc.nextInt(), b = sc.nextInt(), c = sc.nextInt(), t = sc.nextInt();
			if((n|m|a|b|c|t) == 0) break;
			MOD = m;
			long[] s = new long[n];
			for(int i=0;i<n;i++) {
				s[i] = sc.nextLong();
			}
			
			long[][] A = new long[n][n];
			for(int i=0;i<n;i++) {
				if(i-1 >= 0) A[i][i-1] = a;
				A[i][i] = b;
				if(i+1 < n) A[i][i+1] = c;
			}
//			debug(A);
			A = pow_mod(A, t);
			long[] ans = new long[n];
			for(int i=0;i<n;i++) {
				for(int j=0;j<n;j++) ans[i] = ( ans[i] + A[i][j] * s[j]) % MOD;
			}
			
			for(int i=0;i<n;i++) System.out.print(ans[i] + (i==n-1? "\n": " "));
		}
	}
	
	long[][] pow_mod(long[][] A, int t) {
		long[][] ret = new long[n][n];
		for(int i=0;i<n;i++)ret[i][i] = 1;
		
		while(t>0) {
			if((t&1)==1) {
				ret = mlt_mod(ret, A);
			}
			A = mlt_mod(A, A);
			t >>= 1;
//			debug(ret, A, t);
		}
		
		return ret;
	}
	
	long[][] mlt_mod(long[][] A, long[][] B) {
		long[][] ret = new long[n][n];
		for(int i=0;i<n;i++) for(int j=0;j<n;j++) {
			for(int k=0;k<n;k++) ret[i][j] = (ret[i][j] + A[i][k] * B[k][j]) % MOD;
		}
		return ret;
	}
	
	void debug(Object...os) {
		System.err.println(deepToString(os));
	}
	
	public static void main(String[] args ){
		new Main().run();
	}
}