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

typedef vector<int> vec;
typedef vector<vec> mat;//[y][x] == [行][列]
typedef long long ll;
int N,M,A,B,C,T;

void view(const mat& a ){
  cout << "view mat" << endl;
  for(int i=0;i<a.size();i++){
    for(int j=0;j<(int)a[i].size();j++){
      cout << a[i][j] << " ";
    }
    cout << endl;
  }
  cout << endl;
}


int mult(int a,int b,int mod){
  ll x,y,z;
  x = a; y = b; z = mod;
  x = ( x * y ) % z;
  return(int)x;
}

mat matrixmult( const mat &a, const mat &b ,int mod){
  mat ret( a.size(),vec(b[0].size(),0) );
  for(int i=0;i<a.size();i++){
    for(int j=0;j<b[0].size();j++){
      int x = 0;
      for(int k=0;k<min(a[0].size(),b.size());k++){
        x += mult(a[i][k] ,b[k][j], mod);
        x %= mod;
      }
      ret[i][j] = x;
    }
  }
  return ret;
}

mat powm(mat a,mat b,int t){
  //  view(a);
  //view(b);
  while(t>0){
    if( t & 1 ){
      a = matrixmult(a,b,M);
    }
    b = matrixmult(b,b,M);
    t>>=1;
    //view(a);
    // view(b);
  }
  return a;
}


int main(){
  while( cin >> N >>M >>A >> B >> C >> T && (N|M|A|B|C|T) ){
    mat s( 1, vec(N,0) );
    for(int i=0;i<N;i++){
      cin >> s[0][i];
    }
    mat b(N,vec(N,0));
    for(int i=0;i<N;i++){
      if( i-1 >= 0 ) b[i-1][i] = A;
      b[i][i] = B;
      if( i+1 < N ) b[i+1][i] = C;
    }

    mat res = powm(s,b,T);

    for(int i=0;i<res[0].size();i++){
      if( i )  cout << " ";
      cout << res[0][i];
    }
    cout << endl;
  }
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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;

typedef vector<int> vec;
typedef vector<vec> mat;//[y][x] == [行][列]
typedef long long ll;
int N,M,A,B,C,T;

void view(const mat& a ){
  cout << "view mat" << endl;
  for(int i=0;i<a.size();i++){
    for(int j=0;j<(int)a[i].size();j++){
      cout << a[i][j] << " ";
    }
    cout << endl;
  }
  cout << endl;
}


int mult(int a,int b,int mod){
  ll x,y,z;
  x = a; y = b; z = mod;
  x = ( x * y ) % z;
  return(int)x;
}

mat matrixmult( const mat &a, const mat &b ,int mod){
  mat ret( a.size(),vec(b[0].size(),0) );
  for(int i=0;i<a.size();i++){
    for(int j=0;j<b[0].size();j++){
      int x = 0;
      for(int k=0;k<min(a[0].size(),b.size());k++){
        x += mult(a[i][k] ,b[k][j], mod);
        x %= mod;
      }
      ret[i][j] = x;
    }
  }
  return ret;
}

mat powm(mat a,mat b,int t){
  //  view(a);
  //view(b);
  while(t>0){
    if( t & 1 ){
      a = matrixmult(a,b,M);
    }
    b = matrixmult(b,b,M);
    t>>=1;
    //view(a);
    // view(b);
  }
  return a;
}


int main(){
  while( cin >> N >>M >>A >> B >> C >> T && (N|M|A|B|C|T) ){
    mat s( 1, vec(N,0) );
    for(int i=0;i<N;i++){
      cin >> s[0][i];
    }
    mat b(N,vec(N,0));
    for(int i=0;i<N;i++){
      if( i-1 >= 0 ) b[i-1][i] = A;
      b[i][i] = B;
      if( i+1 < N ) b[i+1][i] = C;
    }

    mat res = powm(s,b,T);

    for(int i=0;i<res[0].size();i++){
      if( i )  cout << " ";
      cout << res[0][i];
    }
    cout << endl;
  }
}
```