#include <valarray>
#include <cstdio>
using namespace std;
typedef valarray<int>V;
int n,m,a,b,c,t;
V M(const V &x,const V &y){
	V z(n*n);
	int i=0,j;
	for(;i<n;i++)for(j=0;j<n;j++)z[i*n+j]=(x[slice(i*n,n,1)]*y[slice(j,n,n)]).sum()%m;
	return z;
}
V E(const V &x,const int &n){
	if(n==1)return x;
	if(n&1)return M(E(x,n-1),x);
	V y=E(x,n>>1);
	return M(y,y);
}
V s,x;
void O(const V &r){
	int i=0;
    for(;i<n-1;i++)printf("%d ",(r[slice(i*n,n,1)]*s).sum()%m);
    printf("%d\n",(r[slice(i*n,n,1)]*s).sum()%m);
}
int main(){
	int i;
	for(;scanf("%d%d%d%d%d%d",&n,&m,&a,&b,&c,&t),n;){
		s.resize(n);
		for(i=0;i<n;i++)scanf("%d",&s[i]);
		x.resize(0);
		x.resize(n*n);
		if(t){
			x[0]=b;
			x[1]=c;
			for(i=1;i<n-1;i++){
				x[i*n+i-1]=a;
				x[i*n+i]=b;
				x[i*n+i+1]=c;
			}
			x[i*n+i-1]=a;
			x[i*n+i]=b;
			O(E(x,t));
		}else{
			for(i=0;i<n;i++)x[i*n+i]=1;
			O(x);
		}
	}
}
### Prompt

Your task is to create a Cpp solution to the following 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

### Response

```cpp
#include <valarray>
#include <cstdio>
using namespace std;
typedef valarray<int>V;
int n,m,a,b,c,t;
V M(const V &x,const V &y){
	V z(n*n);
	int i=0,j;
	for(;i<n;i++)for(j=0;j<n;j++)z[i*n+j]=(x[slice(i*n,n,1)]*y[slice(j,n,n)]).sum()%m;
	return z;
}
V E(const V &x,const int &n){
	if(n==1)return x;
	if(n&1)return M(E(x,n-1),x);
	V y=E(x,n>>1);
	return M(y,y);
}
V s,x;
void O(const V &r){
	int i=0;
    for(;i<n-1;i++)printf("%d ",(r[slice(i*n,n,1)]*s).sum()%m);
    printf("%d\n",(r[slice(i*n,n,1)]*s).sum()%m);
}
int main(){
	int i;
	for(;scanf("%d%d%d%d%d%d",&n,&m,&a,&b,&c,&t),n;){
		s.resize(n);
		for(i=0;i<n;i++)scanf("%d",&s[i]);
		x.resize(0);
		x.resize(n*n);
		if(t){
			x[0]=b;
			x[1]=c;
			for(i=1;i<n-1;i++){
				x[i*n+i-1]=a;
				x[i*n+i]=b;
				x[i*n+i+1]=c;
			}
			x[i*n+i-1]=a;
			x[i*n+i]=b;
			O(E(x,t));
		}else{
			for(i=0;i<n;i++)x[i*n+i]=1;
			O(x);
		}
	}
}
```