#include<cstdio>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;

int M;

const int N_MAX=50;

template<class T>
void mul(int n,const T A[N_MAX][N_MAX],const T x[N_MAX],T y[N_MAX]){
	static T z[N_MAX];
	rep(i,n){
		z[i]=0;
		rep(j,n) z[i]=(z[i]+A[i][j]*x[j])%M;
	}
	rep(i,n) y[i]=z[i];
}

template<class T>
void mul(int n,const T A[N_MAX][N_MAX],const T B[N_MAX][N_MAX],T C[N_MAX][N_MAX]){
	static T tmp[N_MAX][N_MAX];
	rep(i,n) rep(j,n) {
		tmp[i][j]=0;
		rep(k,n) tmp[i][j]=(tmp[i][j]+A[i][k]*B[k][j])%M;
	}
	rep(i,n) rep(j,n) C[i][j]=tmp[i][j];
}

template<class T>
void pow(int n,const T A[N_MAX][N_MAX],int m,T B[N_MAX][N_MAX]){
	static T tmp[N_MAX][N_MAX];
	rep(i,n) rep(j,n) {
		tmp[i][j]=A[i][j];
		B[i][j]=(i==j?1:0);
	}
	for(int t=0;m>0;t++){
		if(m&(1LL<<t)){
			mul(n,B,tmp,B);
			m-=1LL<<t;
		}
		mul(n,tmp,tmp,tmp);
	}
}

int main(){
	for(int n,a,b,c,t;scanf("%d%d%d%d%d%d",&n,&M,&a,&b,&c,&t),n;){
		int x[N_MAX];
		rep(i,n) scanf("%d",x+i);

		int A[N_MAX][N_MAX]={};
		rep(i,n){
			if(i> 0 ) A[i][i-1]=a;
			A[i][i]=b;
			if(i<n-1) A[i][i+1]=c;
		}

		pow(n,A,t,A);
		mul(n,A,x,x);

		rep(i,n) printf("%d%c",x[i],i+1<n?' ':'\n');
	}

	return 0;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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<cstdio>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;

int M;

const int N_MAX=50;

template<class T>
void mul(int n,const T A[N_MAX][N_MAX],const T x[N_MAX],T y[N_MAX]){
	static T z[N_MAX];
	rep(i,n){
		z[i]=0;
		rep(j,n) z[i]=(z[i]+A[i][j]*x[j])%M;
	}
	rep(i,n) y[i]=z[i];
}

template<class T>
void mul(int n,const T A[N_MAX][N_MAX],const T B[N_MAX][N_MAX],T C[N_MAX][N_MAX]){
	static T tmp[N_MAX][N_MAX];
	rep(i,n) rep(j,n) {
		tmp[i][j]=0;
		rep(k,n) tmp[i][j]=(tmp[i][j]+A[i][k]*B[k][j])%M;
	}
	rep(i,n) rep(j,n) C[i][j]=tmp[i][j];
}

template<class T>
void pow(int n,const T A[N_MAX][N_MAX],int m,T B[N_MAX][N_MAX]){
	static T tmp[N_MAX][N_MAX];
	rep(i,n) rep(j,n) {
		tmp[i][j]=A[i][j];
		B[i][j]=(i==j?1:0);
	}
	for(int t=0;m>0;t++){
		if(m&(1LL<<t)){
			mul(n,B,tmp,B);
			m-=1LL<<t;
		}
		mul(n,tmp,tmp,tmp);
	}
}

int main(){
	for(int n,a,b,c,t;scanf("%d%d%d%d%d%d",&n,&M,&a,&b,&c,&t),n;){
		int x[N_MAX];
		rep(i,n) scanf("%d",x+i);

		int A[N_MAX][N_MAX]={};
		rep(i,n){
			if(i> 0 ) A[i][i-1]=a;
			A[i][i]=b;
			if(i<n-1) A[i][i+1]=c;
		}

		pow(n,A,t,A);
		mul(n,A,x,x);

		rep(i,n) printf("%d%c",x[i],i+1<n?' ':'\n');
	}

	return 0;
}
```