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
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
typedef vector<int> vec;
typedef vector<vec> mat;
typedef long long ll;

int m;

mat mul(mat &A,mat &B){
	mat C(A[0].size(),vec(B.size()));
	for(int i=0;i<B.size();i++){
		for(int j=0;j<A[0].size();j++){
			for(int k=0;k<B[0].size();k++){
				C[j][i]=(C[j][i]+A[k][i]*B[j][k])%m;
			}
		}
	}
	return C;
}

mat pow(mat A,int n){
	mat B(A.size(),vec(A.size()));
	for(int i=0;i<A.size();i++){
		B[i][i]=1;
	}
	while(n>0){
		if(n & 1)B=mul(B,A);
		A=mul(A,A);
		n>>=1;
	}
	return B;
}

int n,a,b,c,t;
int s[501];
int res[501];


int main(void){
	while(1){
		scanf("%d %d %d %d %d %d",&n,&m,&a,&b,&c,&t);
		if(n+m+a+b+c+t==0)break;
		mat A(n, vec(n));
		for(int i=0;i<n;i++){
			for(int j=-1;j<=1;j++){
				if(j+i>=0 && j+i<n){
					if(j==-1)A[j+i][i]=a;
					if(j==0)A[j+i][i]=b;
					if(j==1)A[j+i][i]=c;
				}
			}
		}
		A=pow(A,t);
		for(int i=0;i<n;i++){
			scanf("%d",&s[i]);
		}
		memset(res,0,sizeof(res));
		for(int i=0;i<n;i++){
			for(int j=0;j<n;j++){
				res[i]=(res[i]+s[j]*A[j][i])%m;
			}
		}
		for(int i=0;i<n;i++){
			printf("%d%c",res[i],i==n-1?'\n':' ');
		}
	}
}