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<stdio.h>
int mat[50][50];
int val[50][50];
int tmp[50][50];
int v[50];
int main(){
	int a,b,c,d,e,f;
	while(scanf("%d%d%d%d%d%d",&a,&b,&c,&d,&e,&f),a){
		for(int i=0;i<a;i++)scanf("%d",v+i);
		for(int i=0;i<a;i++)
			for(int j=0;j<a;j++)
				mat[i][j]=val[i][j]=tmp[i][j]=0;
		for(int i=0;i<a;i++)mat[i][i]=1;
		for(int i=0;i<a;i++){
			if(i)val[i][i-1]=c;
			val[i][i]=d;
			if(i<a-1)val[i][i+1]=e;
		}
		while(f){
			if(f%2){
				for(int i=0;i<a;i++)
					for(int j=0;j<a;j++)
						tmp[i][j]=0;
				for(int i=0;i<a;i++)
					for(int j=0;j<a;j++)
						for(int k=0;k<a;k++)
							tmp[i][k]=(tmp[i][k]+mat[i][j]*val[j][k])%b;
				for(int i=0;i<a;i++)
					for(int j=0;j<a;j++)
						mat[i][j]=tmp[i][j];
			}
			f/=2;
			for(int i=0;i<a;i++)
				for(int j=0;j<a;j++)
					tmp[i][j]=0LL;
			for(int i=0;i<a;i++)
				for(int j=0;j<a;j++)
					for(int k=0;k<a;k++)
						tmp[i][k]=(tmp[i][k]+val[i][j]*val[j][k])%b;
			for(int i=0;i<a;i++)
				for(int j=0;j<a;j++)
					val[i][j]=tmp[i][j];
		}
		for(int i=0;i<a;i++){
			if(i)printf(" ");
			int ret=0;
			for(int j=0;j<a;j++)ret=(ret+mat[i][j]*v[j])%b;
			printf("%d",ret);
		}
		printf("\n");
	}
}