#include<iostream>
#include<algorithm>
using namespace std;
int n,m,a,b,c;long t;
int s[50];
int E[50][50];
int A[50][50];
int main()
{
	while(cin>>n>>m>>a>>b>>c>>t,n)
	{
		for(int i=0;i<n;i++)
		{
			cin>>s[i];
			for(int j=0;j<n;j++)E[i][j]=A[i][j]=0;
			E[i][i]=1;
			A[i][i]=b;
			if(i>0)A[i][i-1]=a;
			if(i<n-1)A[i][i+1]=c;
		}
		for(long i=1;i<=t;i<<=1)
		{
			if(t&i)
			{
				int F[50][50]={};
				for(int j=0;j<n;j++)for(int k=0;k<n;k++)
				{
					for(int l=0;l<n;l++)F[j][k]+=E[j][l]*A[l][k];
				}
				for(int j=0;j<n;j++)for(int k=0;k<n;k++)E[j][k]=F[j][k]%m;
			}
			int F[50][50]={};
			for(int j=0;j<n;j++)for(int k=0;k<n;k++)
			{
				for(int l=0;l<n;l++)F[j][k]+=A[j][l]*A[l][k];
			}
			for(int j=0;j<n;j++)for(int k=0;k<n;k++)A[j][k]=F[j][k]%m;
		}
		for(int i=0;i<n;i++)
		{
			int ans=0;
			for(int j=0;j<n;j++)ans=(ans+s[j]*E[i][j])%m;
			cout<<ans<<(i==n-1?"\n":" ");
		}
	}
}


### 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<iostream>
#include<algorithm>
using namespace std;
int n,m,a,b,c;long t;
int s[50];
int E[50][50];
int A[50][50];
int main()
{
	while(cin>>n>>m>>a>>b>>c>>t,n)
	{
		for(int i=0;i<n;i++)
		{
			cin>>s[i];
			for(int j=0;j<n;j++)E[i][j]=A[i][j]=0;
			E[i][i]=1;
			A[i][i]=b;
			if(i>0)A[i][i-1]=a;
			if(i<n-1)A[i][i+1]=c;
		}
		for(long i=1;i<=t;i<<=1)
		{
			if(t&i)
			{
				int F[50][50]={};
				for(int j=0;j<n;j++)for(int k=0;k<n;k++)
				{
					for(int l=0;l<n;l++)F[j][k]+=E[j][l]*A[l][k];
				}
				for(int j=0;j<n;j++)for(int k=0;k<n;k++)E[j][k]=F[j][k]%m;
			}
			int F[50][50]={};
			for(int j=0;j<n;j++)for(int k=0;k<n;k++)
			{
				for(int l=0;l<n;l++)F[j][k]+=A[j][l]*A[l][k];
			}
			for(int j=0;j<n;j++)for(int k=0;k<n;k++)A[j][k]=F[j][k]%m;
		}
		for(int i=0;i<n;i++)
		{
			int ans=0;
			for(int j=0;j<n;j++)ans=(ans+s[j]*E[i][j])%m;
			cout<<ans<<(i==n-1?"\n":" ");
		}
	}
}


```