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

#define int long long

int MOD;

vector<vector<int>> pro(vector<vector<int>>& a, vector<vector<int>>& b){
  vector<vector<int>> c(a.size(), vector<int>(b.size()));
  
  for(int i = 0; i < a.size(); i++){
    for(int j = 0; j < b[0].size(); j++){
      for(int k = 0; k < b.size(); k++){
        c[i][j] += a[i][k]*b[k][j];
        c[i][j] %= MOD;
      }
    }
  }

  return c;
}

vector<vector<int>> power(vector<vector<int>> a, int t){
  vector<vector<int>> res(a.size(), vector<int>(a.size()));
  
  for(int i = 0; i < a.size(); i++){
    res[i][i] = 1;
  }

  while(t){
    // cout<<t<<endl;
    if(t&1){
      res = pro(res, a);
    }
    a = pro(a,a);
    t >>= 1;
  }

  return res;
}

signed main(){
  
  while(true){
    int n, A, B, C, T;
    vector<vector<int>> mat, ans;

    cin>>n>>MOD>>A>>B>>C>>T;

    if(!n) break;

    ans.resize(n,vector<int>(1,0));
    mat.resize(n,vector<int>(n,0));
    
    for(int i = 0; i < n; i++){
      int S;
      
      cin>>S;
      
      ans[i][0] = S;
    }

    for(int i = 1; i < n-1; i++){
      mat[i][i-1] = A;
      mat[i][i] = B;
      mat[i][i+1] = C;
    }
    
    mat[0][0] = B;
    mat[0][1] = C;
    mat[n-1][n-2] = A;
    mat[n-1][n-1] = B;
    
    // cout<<"power "<<endl;
    
    mat = power(mat, T);
    
    //  cout<<"pro "<<endl;
    ans = pro(mat, ans);
    
    //cout<<"<> "<<endl;
    for(int i = 0; i < n; i++){
      if(i) cout<<" ";
      cout<<ans[i][0];
    }
    cout<<endl;
  }
  
  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;

#define int long long

int MOD;

vector<vector<int>> pro(vector<vector<int>>& a, vector<vector<int>>& b){
  vector<vector<int>> c(a.size(), vector<int>(b.size()));
  
  for(int i = 0; i < a.size(); i++){
    for(int j = 0; j < b[0].size(); j++){
      for(int k = 0; k < b.size(); k++){
        c[i][j] += a[i][k]*b[k][j];
        c[i][j] %= MOD;
      }
    }
  }

  return c;
}

vector<vector<int>> power(vector<vector<int>> a, int t){
  vector<vector<int>> res(a.size(), vector<int>(a.size()));
  
  for(int i = 0; i < a.size(); i++){
    res[i][i] = 1;
  }

  while(t){
    // cout<<t<<endl;
    if(t&1){
      res = pro(res, a);
    }
    a = pro(a,a);
    t >>= 1;
  }

  return res;
}

signed main(){
  
  while(true){
    int n, A, B, C, T;
    vector<vector<int>> mat, ans;

    cin>>n>>MOD>>A>>B>>C>>T;

    if(!n) break;

    ans.resize(n,vector<int>(1,0));
    mat.resize(n,vector<int>(n,0));
    
    for(int i = 0; i < n; i++){
      int S;
      
      cin>>S;
      
      ans[i][0] = S;
    }

    for(int i = 1; i < n-1; i++){
      mat[i][i-1] = A;
      mat[i][i] = B;
      mat[i][i+1] = C;
    }
    
    mat[0][0] = B;
    mat[0][1] = C;
    mat[n-1][n-2] = A;
    mat[n-1][n-1] = B;
    
    // cout<<"power "<<endl;
    
    mat = power(mat, T);
    
    //  cout<<"pro "<<endl;
    ans = pro(mat, ans);
    
    //cout<<"<> "<<endl;
    for(int i = 0; i < n; i++){
      if(i) cout<<" ";
      cout<<ans[i][0];
    }
    cout<<endl;
  }
  
  return 0;
}


```