#include <iostream>
#include <vector>
using namespace std;
int n,m,a,b,c,t;
//int ret_mat[50][50];
//int mat[50][50];
vector<vector<int> > mat_pro(vector<vector<int> >mat,vector<vector<int> >mat2){
  vector<vector<int> > ret(n,vector<int>(n,0));
  int i,j,k,sum;
  for(i=0;i<n;i++){
    for(j=0;j<n;j++){
      sum=0;
      for(k=0;k<n;k++){
        sum+=(mat[i][k]*mat2[k][j])%m;
        sum%=m;
      }
      ret[i][j]=sum;
    }
  }
  return ret;
}
vector<vector<int> > pow_mat(vector<vector<int> >mat,int T){
  vector<vector<int> > ret(n,vector<int>(n,0));
  int i;
  for(i=0;i<n;i++)ret[i][i]=1;//単位行列
  while(T>0){
    if((T&1)==1){
      ret = mat_pro(ret, mat);
    }
    mat = mat_pro(mat , mat);
    T >>= 1;
  }
  return ret;
}
vector<vector<int> > init_mat(){
  vector<vector<int> > mat(n,vector<int>(n,0));
  return mat;
}

vector<vector<int> > set_mat(vector<vector<int> >mat){
  mat[0][0]=b;mat[0][1]=c;
  mat[n-1][n-1]=b;mat[n-1][n-2]=a;
  for(int i=1;i<n-1;i++){
    mat[i][i-1]=a;
    mat[i][i]=b;
    mat[i][i+1]=c;
  }
  return mat;
}
vector<int> Ax(vector<vector<int> > mat,vector<int> v){
  vector<int> s(n,0);
  int i,j,k,sum;
  for(i=0;i<n;i++){
    sum=0;
    for(j=0;j<n;j++){
      sum+=(mat[i][j]*v[j])%m;
      sum%=m;
    }
    s[i]=sum;
  }
  return s;
}
int main(){
  int i;
  while(cin >> n >> m >> a >> b >> c >> t , n){
    vector<int> s(n);
    vector<vector<int> > mat;
    for(i=0;i<n;i++)cin >> s[i];
    mat = init_mat();
    mat = set_mat(mat);
    mat = pow_mat(mat,t);
    s = Ax(mat,s);
    for(i=0;i<n;i++){
      cout << s[i];
      if(i!=n-1)cout << " ";
    }
    cout << endl;
  }
  return 0;
}


### Prompt

Create a solution in CPP for 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;
//int ret_mat[50][50];
//int mat[50][50];
vector<vector<int> > mat_pro(vector<vector<int> >mat,vector<vector<int> >mat2){
  vector<vector<int> > ret(n,vector<int>(n,0));
  int i,j,k,sum;
  for(i=0;i<n;i++){
    for(j=0;j<n;j++){
      sum=0;
      for(k=0;k<n;k++){
        sum+=(mat[i][k]*mat2[k][j])%m;
        sum%=m;
      }
      ret[i][j]=sum;
    }
  }
  return ret;
}
vector<vector<int> > pow_mat(vector<vector<int> >mat,int T){
  vector<vector<int> > ret(n,vector<int>(n,0));
  int i;
  for(i=0;i<n;i++)ret[i][i]=1;//単位行列
  while(T>0){
    if((T&1)==1){
      ret = mat_pro(ret, mat);
    }
    mat = mat_pro(mat , mat);
    T >>= 1;
  }
  return ret;
}
vector<vector<int> > init_mat(){
  vector<vector<int> > mat(n,vector<int>(n,0));
  return mat;
}

vector<vector<int> > set_mat(vector<vector<int> >mat){
  mat[0][0]=b;mat[0][1]=c;
  mat[n-1][n-1]=b;mat[n-1][n-2]=a;
  for(int i=1;i<n-1;i++){
    mat[i][i-1]=a;
    mat[i][i]=b;
    mat[i][i+1]=c;
  }
  return mat;
}
vector<int> Ax(vector<vector<int> > mat,vector<int> v){
  vector<int> s(n,0);
  int i,j,k,sum;
  for(i=0;i<n;i++){
    sum=0;
    for(j=0;j<n;j++){
      sum+=(mat[i][j]*v[j])%m;
      sum%=m;
    }
    s[i]=sum;
  }
  return s;
}
int main(){
  int i;
  while(cin >> n >> m >> a >> b >> c >> t , n){
    vector<int> s(n);
    vector<vector<int> > mat;
    for(i=0;i<n;i++)cin >> s[i];
    mat = init_mat();
    mat = set_mat(mat);
    mat = pow_mat(mat,t);
    s = Ax(mat,s);
    for(i=0;i<n;i++){
      cout << s[i];
      if(i!=n-1)cout << " ";
    }
    cout << endl;
  }
  return 0;
}


```