#include<iostream>
#include<vector>
#include<algorithm>

using namespace std;

vector<vector<int> > mult(vector<vector<int> > mat1,vector<vector<int> > mat2,int m){
    int size = mat1.size();
    vector<vector<int> > mul(size,vector<int>(size,0));
    for(int i=0;i<size;i++){
        for(int j=0;j<size;j++){
            for(int k=0;k<size;k++){
                mul[i][j]+=mat1[i][k]*mat2[k][j];
                mul[i][j]%=m;
            }
        }
    }
    return mul;
}

vector<vector<int> > repeat_pow(vector<vector<int> > mat,int t, int m){
    if(t==0){
        int size = mat.size();
        vector<vector<int> > ide(size,vector<int>(size,0));
        for(int i=0;i<size;i++){
            ide[i][i]=1;
        }
        return ide;
    }
    if(t%2==0){
        mat = repeat_pow(mat,t/2,m);
        return mult(mat,mat,m); 
    }else{
        return mult(mat,repeat_pow(mat,t-1,m),m);
    }
}

int main(){
    int n,m,a,b,c,t;
    while(cin >> n >> m >> a >> b >> c >> t,n|m|a|b|c|t){
        vector<int> s(n);
        for(int i=0;i<n;i++){
            cin >> s[i];
        }
        vector<vector<int> > mat(n,vector<int>(n,0));
        for(int i=0;i<n;i++){
            mat[i][i]=b;
            if(i+1<n)mat[i][i+1]=c;
            if(i-1>=0)mat[i][i-1]=a;
        }
        vector<vector<int> > mat_pow(n,vector<int>(n));
        mat_pow = repeat_pow(mat,t,m);
        vector<int> ans(n);
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                ans[i] += s[j] * mat_pow[i][j];
                ans[i]%=m;
            }
        }
        for(int i=0;i<n;i++){
            if(i==0)cout << ans[i];
            else cout << " " << ans[i];
        }
        cout << endl;
    }
    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<vector>
#include<algorithm>

using namespace std;

vector<vector<int> > mult(vector<vector<int> > mat1,vector<vector<int> > mat2,int m){
    int size = mat1.size();
    vector<vector<int> > mul(size,vector<int>(size,0));
    for(int i=0;i<size;i++){
        for(int j=0;j<size;j++){
            for(int k=0;k<size;k++){
                mul[i][j]+=mat1[i][k]*mat2[k][j];
                mul[i][j]%=m;
            }
        }
    }
    return mul;
}

vector<vector<int> > repeat_pow(vector<vector<int> > mat,int t, int m){
    if(t==0){
        int size = mat.size();
        vector<vector<int> > ide(size,vector<int>(size,0));
        for(int i=0;i<size;i++){
            ide[i][i]=1;
        }
        return ide;
    }
    if(t%2==0){
        mat = repeat_pow(mat,t/2,m);
        return mult(mat,mat,m); 
    }else{
        return mult(mat,repeat_pow(mat,t-1,m),m);
    }
}

int main(){
    int n,m,a,b,c,t;
    while(cin >> n >> m >> a >> b >> c >> t,n|m|a|b|c|t){
        vector<int> s(n);
        for(int i=0;i<n;i++){
            cin >> s[i];
        }
        vector<vector<int> > mat(n,vector<int>(n,0));
        for(int i=0;i<n;i++){
            mat[i][i]=b;
            if(i+1<n)mat[i][i+1]=c;
            if(i-1>=0)mat[i][i-1]=a;
        }
        vector<vector<int> > mat_pow(n,vector<int>(n));
        mat_pow = repeat_pow(mat,t,m);
        vector<int> ans(n);
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                ans[i] += s[j] * mat_pow[i][j];
                ans[i]%=m;
            }
        }
        for(int i=0;i<n;i++){
            if(i==0)cout << ans[i];
            else cout << " " << ans[i];
        }
        cout << endl;
    }
    return 0;
}

```