#include<iostream>
#include<cstdio>
using namespace std;
#define MAXN 55
#define MOD m
int m;
struct Matrix
{
    int data[MAXN][MAXN];
    int n;
    void init()
    {
        for(int i=0;i<MAXN;i++)
            for(int j=0;j<MAXN;j++)
            data[i][j]=0;
        n=0;
    }
};

Matrix Multy(Matrix m1,Matrix m2)
{
    Matrix r;
    r.init();
    int n=m1.n;
    r.n=n;
    for(int i=0;i<n;i++)
        for(int j=0;j<n;j++)
            for(int k=0;k<n;k++)
            r.data[i][j]=(r.data[i][j]+m1.data[i][k]*m2.data[k][j])%MOD;
    return r;
}

Matrix Fast_mi(Matrix ma,int m)
{
    Matrix r;
    r.init();
    r.n=ma.n;
    int n=ma.n;
    for(int i=0;i<n;i++)
        r.data[i][i]=1;
    while(m)
    {
        if(m&1)r=Multy(r,ma);
        ma=Multy(ma,ma);
        m=m>>1;
    }
    return r;
}

int Tr(Matrix ma)
{
    int sum=0;
    for(int i=0;i<ma.n;i++)
        sum=(sum+ma.data[i][i])%MOD;
    return sum;
}

int main()
{
    int n,a,b,c,t;
    while(scanf("%d%d%d%d%d%d",&n,&m,&a,&b,&c,&t)!=EOF)
    {
        if(n==0&&m==n&&a==n&&b==n&&c==n&&t==n)break;
        Matrix ma,s;
        ma.init();s.init();
        ma.n=n;s.n=n;
        for(int i=0;i<n;++i)
            scanf("%d",&s.data[0][i]);
        for(int i=0;i<n;i++)
        {
            if(i-1>=0)ma.data[i-1][i]=a;
            ma.data[i][i]=b;
            if(i+1<n)ma.data[i+1][i]=c;
        }
        s=Multy(s,Fast_mi(ma,t));
        printf("%d",s.data[0][0]);
        for(int i=1;i<n;++i)
            printf(" %d",s.data[0][i]);
        printf("\n");
    }
    return 0;
}
### Prompt

In Cpp, your task is to solve 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<cstdio>
using namespace std;
#define MAXN 55
#define MOD m
int m;
struct Matrix
{
    int data[MAXN][MAXN];
    int n;
    void init()
    {
        for(int i=0;i<MAXN;i++)
            for(int j=0;j<MAXN;j++)
            data[i][j]=0;
        n=0;
    }
};

Matrix Multy(Matrix m1,Matrix m2)
{
    Matrix r;
    r.init();
    int n=m1.n;
    r.n=n;
    for(int i=0;i<n;i++)
        for(int j=0;j<n;j++)
            for(int k=0;k<n;k++)
            r.data[i][j]=(r.data[i][j]+m1.data[i][k]*m2.data[k][j])%MOD;
    return r;
}

Matrix Fast_mi(Matrix ma,int m)
{
    Matrix r;
    r.init();
    r.n=ma.n;
    int n=ma.n;
    for(int i=0;i<n;i++)
        r.data[i][i]=1;
    while(m)
    {
        if(m&1)r=Multy(r,ma);
        ma=Multy(ma,ma);
        m=m>>1;
    }
    return r;
}

int Tr(Matrix ma)
{
    int sum=0;
    for(int i=0;i<ma.n;i++)
        sum=(sum+ma.data[i][i])%MOD;
    return sum;
}

int main()
{
    int n,a,b,c,t;
    while(scanf("%d%d%d%d%d%d",&n,&m,&a,&b,&c,&t)!=EOF)
    {
        if(n==0&&m==n&&a==n&&b==n&&c==n&&t==n)break;
        Matrix ma,s;
        ma.init();s.init();
        ma.n=n;s.n=n;
        for(int i=0;i<n;++i)
            scanf("%d",&s.data[0][i]);
        for(int i=0;i<n;i++)
        {
            if(i-1>=0)ma.data[i-1][i]=a;
            ma.data[i][i]=b;
            if(i+1<n)ma.data[i+1][i]=c;
        }
        s=Multy(s,Fast_mi(ma,t));
        printf("%d",s.data[0][0]);
        for(int i=1;i<n;++i)
            printf(" %d",s.data[0][i]);
        printf("\n");
    }
    return 0;
}
```