#include<cstdio>
#include<cstring>
const int maxn=55;
int n,m,a,b,c,t;
int s[maxn];
struct matrix
{
    int mat[maxn][maxn];
};
matrix operator * (const matrix &A,const matrix &B)
{
    matrix C;
    memset(C.mat,0,sizeof(C.mat));
    for (int k=0;k<n;k++)
        for (int i=0;i<n;i++)
            if (A.mat[i][k])
                for (int j=0;j<n;j++)
                    C.mat[i][j]=(C.mat[i][j]+A.mat[i][k]*B.mat[k][j])%m;
    return C;
}
matrix v_pow(const matrix &A,int nn)
{
    matrix B,C;
    memset(C.mat,0,sizeof(C.mat));
    for (int i=0;i<n;i++) C.mat[i][i]=1;
    B=A;
    while (nn)
    {
        if (nn&1) C=C*B;
        B=B*B;
        nn>>=1;
    }
    return C;
}
int main()
{
    while (~scanf("%d%d%d%d%d%d",&n,&m,&a,&b,&c,&t)&&(n||m||a||b||c||t))
    {
        for (int i=0;i<n;i++) scanf("%d",&s[i]);
        matrix A;
        memset(A.mat,0,sizeof(A.mat));
        for (int i=0;i<n;i++)
        {
            if (i) A.mat[i][i-1]=a;
            A.mat[i][i]=b;
            if (i!=n-1) A.mat[i][i+1]=c;
        }
        A=v_pow(A,t);
        int f=0;
        for (int i=0;i<n;i++)
        {
            int ans=0;
            for (int j=0;j<n;j++) ans=(ans+A.mat[i][j]*s[j])%m;
            if (f) putchar(' ');
            else f=1;
            printf("%d",ans);
        }
        putchar('\n');
    }
    return 0;
}
### 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<cstdio>
#include<cstring>
const int maxn=55;
int n,m,a,b,c,t;
int s[maxn];
struct matrix
{
    int mat[maxn][maxn];
};
matrix operator * (const matrix &A,const matrix &B)
{
    matrix C;
    memset(C.mat,0,sizeof(C.mat));
    for (int k=0;k<n;k++)
        for (int i=0;i<n;i++)
            if (A.mat[i][k])
                for (int j=0;j<n;j++)
                    C.mat[i][j]=(C.mat[i][j]+A.mat[i][k]*B.mat[k][j])%m;
    return C;
}
matrix v_pow(const matrix &A,int nn)
{
    matrix B,C;
    memset(C.mat,0,sizeof(C.mat));
    for (int i=0;i<n;i++) C.mat[i][i]=1;
    B=A;
    while (nn)
    {
        if (nn&1) C=C*B;
        B=B*B;
        nn>>=1;
    }
    return C;
}
int main()
{
    while (~scanf("%d%d%d%d%d%d",&n,&m,&a,&b,&c,&t)&&(n||m||a||b||c||t))
    {
        for (int i=0;i<n;i++) scanf("%d",&s[i]);
        matrix A;
        memset(A.mat,0,sizeof(A.mat));
        for (int i=0;i<n;i++)
        {
            if (i) A.mat[i][i-1]=a;
            A.mat[i][i]=b;
            if (i!=n-1) A.mat[i][i+1]=c;
        }
        A=v_pow(A,t);
        int f=0;
        for (int i=0;i<n;i++)
        {
            int ans=0;
            for (int j=0;j<n;j++) ans=(ans+A.mat[i][j]*s[j])%m;
            if (f) putchar(' ');
            else f=1;
            printf("%d",ans);
        }
        putchar('\n');
    }
    return 0;
}
```