#include <bits/stdc++.h>
using namespace std;

int N, M, A, B, C, T;
vector<int> S;

typedef vector<int> vec;
typedef vector<vec> mat;

mat mul(mat &A, mat &B){
    mat C(A.size(), vec(B[0].size()));
    for(int i=0; i<A.size(); i++){
        for(int k=0; k<B.size(); k++){
            for(int j=0; j<B[0].size(); j++){
                C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % M;
            }
        }
    }
    return C;
}

mat pow(mat A, int n){
    mat B(A.size(), vec(A.size()));
    for(int i=0; i<A.size(); i++){
        B[i][i] = 1;
    }
    while(n>0){
        if(n&1) B = mul(B, A);
        A = mul(A, A);
        n >>= 1;
    }
    return B;
}


int main(){
    while(cin>>N>>M>>A>>B>>C>>T, N|M|A|B|C|T){
        S.resize(N);
        for(auto &s: S) cin >> s;
        
        vector<int> co({A, B, C});
        mat Mat = mat(N, vec(N));
        for(int i=0; i<N; i++){
            for(int j=0; j<3; j++) if(0<=j+i-1 && j+i-1<N){
                Mat[i][j+i-1] = co[j];
            }
        }
        
        mat init;
        for(int i=0; i<N; i++)
            init.push_back(vec({S[i]}));
        Mat = pow(Mat, T);
        Mat = mul(Mat, init);
        
        
        for(int i=0; i<N; i++)
            cout << Mat[i][0] << (i==N-1? "\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 <bits/stdc++.h>
using namespace std;

int N, M, A, B, C, T;
vector<int> S;

typedef vector<int> vec;
typedef vector<vec> mat;

mat mul(mat &A, mat &B){
    mat C(A.size(), vec(B[0].size()));
    for(int i=0; i<A.size(); i++){
        for(int k=0; k<B.size(); k++){
            for(int j=0; j<B[0].size(); j++){
                C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % M;
            }
        }
    }
    return C;
}

mat pow(mat A, int n){
    mat B(A.size(), vec(A.size()));
    for(int i=0; i<A.size(); i++){
        B[i][i] = 1;
    }
    while(n>0){
        if(n&1) B = mul(B, A);
        A = mul(A, A);
        n >>= 1;
    }
    return B;
}


int main(){
    while(cin>>N>>M>>A>>B>>C>>T, N|M|A|B|C|T){
        S.resize(N);
        for(auto &s: S) cin >> s;
        
        vector<int> co({A, B, C});
        mat Mat = mat(N, vec(N));
        for(int i=0; i<N; i++){
            for(int j=0; j<3; j++) if(0<=j+i-1 && j+i-1<N){
                Mat[i][j+i-1] = co[j];
            }
        }
        
        mat init;
        for(int i=0; i<N; i++)
            init.push_back(vec({S[i]}));
        Mat = pow(Mat, T);
        Mat = mul(Mat, init);
        
        
        for(int i=0; i<N; i++)
            cout << Mat[i][0] << (i==N-1? "\n": " ");
    }
    
    return 0;
}
```