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 CPP):

Solution:
#include <iostream>
#include <sstream>
#include <vector>
#include <string>
#include <algorithm>
#include <utility>
#include <set>
#include <map>
#include <deque>
#include <queue>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cstdio>
#include <stdio.h>

using namespace std;

#define fo(i,n) for(int i=0; i<(int)n; i++)
#define rep(it,s) for(__typeof((s).begin()) it=(s).begin();it!=(s).end();it++)
#define mp(a,b) make_pair(a,b)
#define pb(x) push_back(x)
#define pii pair<int,int>

#define vll vector<long long>
#define vvll vector< vll >

int N,M,A,B,C,T;

int s[55];

vvll mul(vvll a, vvll b) {
	vvll c = vvll(N,vll(N,0)); 
	
	for (int i=0; i<N; i++) for (int j=0; j<N; j++) {
		long long sum = 0;
		for (int k=0; k<N; k++) {
			sum += a[i][k]*b[k][j]%M;
			sum %= M;
		}
		c[i][j] = sum;
	}
	return c;
}

vvll mpow(vvll a, int e) {
	if (e==1) return a;
	
	vvll c = vvll(N,vll(N,0)); 
	for (int i=0; i<N; i++) {
		c[i][i] = 1;
	}
	
	if (e==0) return c;
	
	c = mpow(a,e>>1);
	c = mul(c,c);
	if (e%2==0) return c;
	return mul(a,c);
}
	

int main() {

    //freopen("input.txt","r",stdin);
    //freopen("output.txt","w",stdout);

	while (cin>>N>>M>>A>>B>>C>>T && N!=0) {
		for (int i=0; i<N; i++) cin>>s[i];
		
		vvll trans = vvll(N,vll(N,0));
		
		for (int i=0; i<N; i++) {
			trans[i][i] = B;
			if (i>0) trans[i][i-1] = A;
			if (i<N-1) trans[i][i+1] = C;
		}
		
		trans = mpow(trans,T);
		
		for (int i=0; i<N; i++) {
			long long sum = 0;
			for (int k=0; k<N; k++) {
				sum += (s[k]*trans[i][k])%M;
				sum %= M;
			}
			cout<<sum;
			if (i!=N-1) cout<<" ";
		}
		
		cout<<endl;
	}
	
    return 0;

}