#include<iostream>
#include<string>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<stack>
#include<queue>
#include<cmath>
#include<numeric>
#include<cassert>
#include<limits>
#include<iomanip>
using namespace std;
typedef vector<int> vi;
typedef vector<vi> vvi;


vvi search(int n,int m,int a,int b,int c,int t){
    vvi r(n);
    for(int i=0;i<n;i++){
        r[i].resize(n);
    }
    if(t==0){
        for(int i=0;i<n;i++){
            r[i][i]=1;
        }
        return r;
    }
    if(t==1){
        for(int i=0;i<n;i++){
            if(i-1>=0){r[i-1][i]=a%m;}
            r[i][i]=b%m;
            if(i+1<n){r[i+1][i]=c%m;}
        }
        return r;
    }
    vvi ret=search(n,m,a,b,c,t/2);
    for(int i=0;i<n;i++){
        for(int h=0;h<n;h++){
            for(int k=0;k<n;k++){
                r[k][h]+=ret[i][h]*ret[k][i];
                r[k][h]%=m;
            }
        }
    }
    if(t%2==1){
        for(int i=0;i<ret.size();i++){
            for(int h=0;h<ret[i].size();h++){
                ret[i][h]=0;
            }
        }
        for(int i=0;i<n;i++){
            for(int h=0;h<n;h++){
                if(i-1>=0){ret[i-1][h]+=r[i][h]*a%m; ret[i-1][h]%=m;}
                ret[i][h]+=r[i][h]*b%m; ret[i][h]%=m;
                if(i+1<n){ret[i+1][h]+=r[i][h]*c; ret[i+1][h]%=m;}
            }
        }
        return ret;
    }
    return r;
}


int main(){
    while(1){
        
        
    int n,m,a,b,c,t;
    cin>>n>>m>>a>>b>>c>>t;
        if(n==0 && m==0 && a==0 && b==0 && c==0 && t==0){break;}
    swap(a,c);
    vi ans(n);
    vi in(n);
    for(int i=0;i<n;i++){
        cin>>in[i];
    }
    vvi vv=search(n,m,a,b,c,t);
    for(int i=0;i<n;i++){
        for(int h=0;h<n;h++){
            ans[i]+=in[h]*vv[i][h]%m;
        }
    }
    for(int i=0;i<n;i++){
        ans[i]%=m;
        cout<<ans[i];
        if(i!=n-1){cout<<" ";}
    }
    cout<<endl;
        
        
    }
    return 0;
}


### Prompt

Develop a solution in CPP to the problem described below:

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<string>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<stack>
#include<queue>
#include<cmath>
#include<numeric>
#include<cassert>
#include<limits>
#include<iomanip>
using namespace std;
typedef vector<int> vi;
typedef vector<vi> vvi;


vvi search(int n,int m,int a,int b,int c,int t){
    vvi r(n);
    for(int i=0;i<n;i++){
        r[i].resize(n);
    }
    if(t==0){
        for(int i=0;i<n;i++){
            r[i][i]=1;
        }
        return r;
    }
    if(t==1){
        for(int i=0;i<n;i++){
            if(i-1>=0){r[i-1][i]=a%m;}
            r[i][i]=b%m;
            if(i+1<n){r[i+1][i]=c%m;}
        }
        return r;
    }
    vvi ret=search(n,m,a,b,c,t/2);
    for(int i=0;i<n;i++){
        for(int h=0;h<n;h++){
            for(int k=0;k<n;k++){
                r[k][h]+=ret[i][h]*ret[k][i];
                r[k][h]%=m;
            }
        }
    }
    if(t%2==1){
        for(int i=0;i<ret.size();i++){
            for(int h=0;h<ret[i].size();h++){
                ret[i][h]=0;
            }
        }
        for(int i=0;i<n;i++){
            for(int h=0;h<n;h++){
                if(i-1>=0){ret[i-1][h]+=r[i][h]*a%m; ret[i-1][h]%=m;}
                ret[i][h]+=r[i][h]*b%m; ret[i][h]%=m;
                if(i+1<n){ret[i+1][h]+=r[i][h]*c; ret[i+1][h]%=m;}
            }
        }
        return ret;
    }
    return r;
}


int main(){
    while(1){
        
        
    int n,m,a,b,c,t;
    cin>>n>>m>>a>>b>>c>>t;
        if(n==0 && m==0 && a==0 && b==0 && c==0 && t==0){break;}
    swap(a,c);
    vi ans(n);
    vi in(n);
    for(int i=0;i<n;i++){
        cin>>in[i];
    }
    vvi vv=search(n,m,a,b,c,t);
    for(int i=0;i<n;i++){
        for(int h=0;h<n;h++){
            ans[i]+=in[h]*vv[i][h]%m;
        }
    }
    for(int i=0;i<n;i++){
        ans[i]%=m;
        cout<<ans[i];
        if(i!=n-1){cout<<" ";}
    }
    cout<<endl;
        
        
    }
    return 0;
}


```