#include <iostream>
#include <vector>
using namespace std;

int N,M,A,B,C,T;
struct matrix{
    vector<vector<int>> m;
    matrix(int N){
        m = vector<vector<int>>(N,vector<int>(N,0));
    }
};

matrix mult(matrix& A,matrix& B){
    matrix C(N);
    for(int i=0;i<N;i++) for(int j=0;j<N;j++) for(int k=0;k<N;k++)
    (C.m[i][j] += A.m[i][k]*B.m[k][j])%=M;
    return C;
}

matrix pow(matrix& A,int n){
    if(n==1) return A;
    if(n&1){
        matrix P = pow(A,n-1);
        return mult(P,A);
    }else{
        matrix P = pow(A,n>>1);
        return mult(P,P);        
    }
}

void solve(vector<int>& S){
    if(T==0){
        for(int i=0;i<N;i++) cout << S[i] << (i!=N-1? " ":"\n");
        return ;
    }
    matrix X(N);
    for(int i=0;i<N;i++){
        if(i-1>=0) X.m[i][i-1] = A;
        X.m[i][i] = B;
        if(i+1<N) X.m[i][i+1] = C;
    }
    matrix res = pow(X,T);
    for(int i=0;i<N;i++){
        int ans = 0;
        for(int j=0;j<N;j++){
            (ans += res.m[i][j]*S[j])%=M;
        }
        cout << ans << (i!=N-1? " ":"\n");
    }
}

int main(){
    while(cin >> N >> M >> A >> B >> C >> T && N){
        vector<int> S(N);
        for(int i=0;i<N;i++) cin >> S[i];
        solve(S);
    }
}

### Prompt

Please formulate 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 <iostream>
#include <vector>
using namespace std;

int N,M,A,B,C,T;
struct matrix{
    vector<vector<int>> m;
    matrix(int N){
        m = vector<vector<int>>(N,vector<int>(N,0));
    }
};

matrix mult(matrix& A,matrix& B){
    matrix C(N);
    for(int i=0;i<N;i++) for(int j=0;j<N;j++) for(int k=0;k<N;k++)
    (C.m[i][j] += A.m[i][k]*B.m[k][j])%=M;
    return C;
}

matrix pow(matrix& A,int n){
    if(n==1) return A;
    if(n&1){
        matrix P = pow(A,n-1);
        return mult(P,A);
    }else{
        matrix P = pow(A,n>>1);
        return mult(P,P);        
    }
}

void solve(vector<int>& S){
    if(T==0){
        for(int i=0;i<N;i++) cout << S[i] << (i!=N-1? " ":"\n");
        return ;
    }
    matrix X(N);
    for(int i=0;i<N;i++){
        if(i-1>=0) X.m[i][i-1] = A;
        X.m[i][i] = B;
        if(i+1<N) X.m[i][i+1] = C;
    }
    matrix res = pow(X,T);
    for(int i=0;i<N;i++){
        int ans = 0;
        for(int j=0;j<N;j++){
            (ans += res.m[i][j]*S[j])%=M;
        }
        cout << ans << (i!=N-1? " ":"\n");
    }
}

int main(){
    while(cin >> N >> M >> A >> B >> C >> T && N){
        vector<int> S(N);
        for(int i=0;i<N;i++) cin >> S[i];
        solve(S);
    }
}

```