#include <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<n;i++)
#define INF 100000000
using namespace std;

int n,m,a,b,c,t;
int x[50];
int mat[50][50],ans[50][50];

void A(int mx[50][50],int my[50][50],int check){
	int z[50][50];
	memset(z,0,sizeof(z));
	rep(i,n)rep(j,n)rep(k,n){
		z[i][j]+=mx[i][k]*my[k][j];
		z[i][j]%=m;
	}
	if(check)rep(i,n)rep(j,n)ans[i][j]=z[i][j];
	else rep(i,n)rep(j,n)mat[i][j]=z[i][j];
}

int main(){
	while(cin>>n>>m>>a>>b>>c>>t,n||m||a||b||c||t){
		rep(i,n)cin>>x[i];
		memset(ans,0,sizeof(ans));
		memset(mat,0,sizeof(mat));
		rep(i,n)ans[i][i]=1;
		if(n==1){
			mat[0][0]=b;
		} else{
			mat[0][0]=b;mat[0][1]=c;
			for(int i=1;i<n-1;i++){
				mat[i][i-1]=a;mat[i][i]=b;mat[i][i+1]=c;
			}
			mat[n-1][n-2]=a;mat[n-1][n-1]=b;
		}
		while(t){
			if(t%2) A(ans,mat,1);
			A(mat,mat,0);
			t/=2;
		}
		rep(i,n){
			int out=0;
			rep(j,n)out+=ans[i][j]*x[j];
			if(i)cout<<" ";
			cout<<out%m;
		}
		cout<<endl;
	}
}
### Prompt

Your challenge is to write 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 <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<n;i++)
#define INF 100000000
using namespace std;

int n,m,a,b,c,t;
int x[50];
int mat[50][50],ans[50][50];

void A(int mx[50][50],int my[50][50],int check){
	int z[50][50];
	memset(z,0,sizeof(z));
	rep(i,n)rep(j,n)rep(k,n){
		z[i][j]+=mx[i][k]*my[k][j];
		z[i][j]%=m;
	}
	if(check)rep(i,n)rep(j,n)ans[i][j]=z[i][j];
	else rep(i,n)rep(j,n)mat[i][j]=z[i][j];
}

int main(){
	while(cin>>n>>m>>a>>b>>c>>t,n||m||a||b||c||t){
		rep(i,n)cin>>x[i];
		memset(ans,0,sizeof(ans));
		memset(mat,0,sizeof(mat));
		rep(i,n)ans[i][i]=1;
		if(n==1){
			mat[0][0]=b;
		} else{
			mat[0][0]=b;mat[0][1]=c;
			for(int i=1;i<n-1;i++){
				mat[i][i-1]=a;mat[i][i]=b;mat[i][i+1]=c;
			}
			mat[n-1][n-2]=a;mat[n-1][n-1]=b;
		}
		while(t){
			if(t%2) A(ans,mat,1);
			A(mat,mat,0);
			t/=2;
		}
		rep(i,n){
			int out=0;
			rep(j,n)out+=ans[i][j]*x[j];
			if(i)cout<<" ";
			cout<<out%m;
		}
		cout<<endl;
	}
}
```