#define N (1<<6)
#include <bits/stdc++.h>

using namespace std;

int g[N][N],p[N];
int n,M,a,b,c,t;

struct matrix
{
	int n,a[N][N];
	matrix(int tn,int x):n(tn)
	{
		for(int i=0;i<n;i++)
			for(int j=0;j<n;j++)
				a[i][j]=i==j?x:0;
	}

	matrix(int tn,int g[][N]):n(tn)
	{
		for(int i=0;i<n;i++)
			for(int j=0;j<n;j++)
				a[i][j]=g[i][j];
	}

	matrix operator*(matrix& b)
	{
		matrix c(n,0);
		for(int i=0;i<n;i++)
			for(int j=0;j<n;j++)
				for(int k=0;k<n;k++)
					c.a[i][j]+=a[i][k]*b.a[k][j],c.a[i][j]%=M;
		return c;
	}

	matrix operator^(int k)
	{
		matrix res(n,1),cur(*this);
		for(;k;k>>=1,cur=cur*cur)
			if(k&1) res=res*cur;
		return res;
	}
};

int main()
{
	for(;cin>>n>>M>>a>>b>>c>>t;)
	{
		memset(g,0,sizeof(g));
		for(int i=0;i<n;i++)
			scanf("%d",p+i);
		for(int i=0;i<n;i++)
		{
			g[i][i]=b;
			if(i>0) g[i-1][i]=a;
			if(i<n-1) g[i+1][i]=c;
		}

		matrix m(n,g); m=m^t;

		for(int i=0;i<n;i++)
		{
			int res=0;
			for(int j=0;j<n;j++) 
				res+=p[j]*m.a[j][i],res%=M;
			if(i<n-1)
				printf("%d ",res);
			else
				printf("%d\n",res);
		}
	}
}
### 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
#define N (1<<6)
#include <bits/stdc++.h>

using namespace std;

int g[N][N],p[N];
int n,M,a,b,c,t;

struct matrix
{
	int n,a[N][N];
	matrix(int tn,int x):n(tn)
	{
		for(int i=0;i<n;i++)
			for(int j=0;j<n;j++)
				a[i][j]=i==j?x:0;
	}

	matrix(int tn,int g[][N]):n(tn)
	{
		for(int i=0;i<n;i++)
			for(int j=0;j<n;j++)
				a[i][j]=g[i][j];
	}

	matrix operator*(matrix& b)
	{
		matrix c(n,0);
		for(int i=0;i<n;i++)
			for(int j=0;j<n;j++)
				for(int k=0;k<n;k++)
					c.a[i][j]+=a[i][k]*b.a[k][j],c.a[i][j]%=M;
		return c;
	}

	matrix operator^(int k)
	{
		matrix res(n,1),cur(*this);
		for(;k;k>>=1,cur=cur*cur)
			if(k&1) res=res*cur;
		return res;
	}
};

int main()
{
	for(;cin>>n>>M>>a>>b>>c>>t;)
	{
		memset(g,0,sizeof(g));
		for(int i=0;i<n;i++)
			scanf("%d",p+i);
		for(int i=0;i<n;i++)
		{
			g[i][i]=b;
			if(i>0) g[i-1][i]=a;
			if(i<n-1) g[i+1][i]=c;
		}

		matrix m(n,g); m=m^t;

		for(int i=0;i<n;i++)
		{
			int res=0;
			for(int j=0;j<n;j++) 
				res+=p[j]*m.a[j][i],res%=M;
			if(i<n-1)
				printf("%d ",res);
			else
				printf("%d\n",res);
		}
	}
}
```