#include<bits/stdc++.h>
using namespace std;
typedef long long lli;
typedef vector< lli > vec;
typedef vector< vec > mat;
 
/*
lli modmul(lli a,lli b,lli m){
  lli r=0;
  for(;b;b>>=1,a=(a<<1)%m)
    if(b&1)
      r=(r+a)%m;
  return r;
}
*/
lli modmul(lli a,lli b,lli m){
  return (((a*(b>>20)%m)<<20)+a*(b&((1<<20)-1)))%m;
}
 
mat matmul(mat &a,mat &b,lli m){
  mat res(a.size(),vec(b[0].size()));
  for(int i=0;i<(int)res.size();i++){
    for(int j=0;j<(int)res[i].size();j++){
      res[i][j]=0;
      for(int k=0;k<(int)b.size();k++){
    res[i][j]+= modmul(a[i][k],b[k][j],m);
    res[i][j]%=m;
      }
    }
  }
  return res;
}
 
mat matpow(mat &a,lli n,lli m){
  mat res(a.size(),vec(a.size()));
  mat x(a.size(),vec(a.size()));
 
  for(int i=0;i<(int)res.size();i++)
    for(int j=0;j<(int)res.size();j++)
      res[i][j]=(i==j?1:0);
 
  for(x=a ; n ; n>>=1,x=matmul(x,x,m) )
    if(n&1)res=matmul(res,x,m);
  return res;
}
 
 
 
 
lli N,M,A,B,C,T;
 
void solve(){
  mat base(N,vec(N));
  for(int i=0;i<N;i++){
    for(int j=0;j<N;j++)base[i][j]=0;
    if(i>0)base[i][i-1]=A%M;
    base[i][i]=B%M;
    if(i+1<N)base[i][i+1]=C%M;
  }
  base = matpow(base,T,M);
  mat ans(N,vec(1));
  for(int i=0;i<N;i++)cin>>ans[i][0];
  ans=matmul(base,ans,M);
  for(int i=0;i<N;i++){
    if(i)cout<<' ';
    cout<<ans[i][0];
  }
  cout<<endl;
}
 
int main(){
  while(cin>>N>>M>>A>>B>>C>>T){
    if(N==0&&M==0&&A==0&&B==0&&C==0&&T==0)break;
    solve();
  }
  return 0;
}
### Prompt

Please create a solution in CPP 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;
typedef long long lli;
typedef vector< lli > vec;
typedef vector< vec > mat;
 
/*
lli modmul(lli a,lli b,lli m){
  lli r=0;
  for(;b;b>>=1,a=(a<<1)%m)
    if(b&1)
      r=(r+a)%m;
  return r;
}
*/
lli modmul(lli a,lli b,lli m){
  return (((a*(b>>20)%m)<<20)+a*(b&((1<<20)-1)))%m;
}
 
mat matmul(mat &a,mat &b,lli m){
  mat res(a.size(),vec(b[0].size()));
  for(int i=0;i<(int)res.size();i++){
    for(int j=0;j<(int)res[i].size();j++){
      res[i][j]=0;
      for(int k=0;k<(int)b.size();k++){
    res[i][j]+= modmul(a[i][k],b[k][j],m);
    res[i][j]%=m;
      }
    }
  }
  return res;
}
 
mat matpow(mat &a,lli n,lli m){
  mat res(a.size(),vec(a.size()));
  mat x(a.size(),vec(a.size()));
 
  for(int i=0;i<(int)res.size();i++)
    for(int j=0;j<(int)res.size();j++)
      res[i][j]=(i==j?1:0);
 
  for(x=a ; n ; n>>=1,x=matmul(x,x,m) )
    if(n&1)res=matmul(res,x,m);
  return res;
}
 
 
 
 
lli N,M,A,B,C,T;
 
void solve(){
  mat base(N,vec(N));
  for(int i=0;i<N;i++){
    for(int j=0;j<N;j++)base[i][j]=0;
    if(i>0)base[i][i-1]=A%M;
    base[i][i]=B%M;
    if(i+1<N)base[i][i+1]=C%M;
  }
  base = matpow(base,T,M);
  mat ans(N,vec(1));
  for(int i=0;i<N;i++)cin>>ans[i][0];
  ans=matmul(base,ans,M);
  for(int i=0;i<N;i++){
    if(i)cout<<' ';
    cout<<ans[i][0];
  }
  cout<<endl;
}
 
int main(){
  while(cin>>N>>M>>A>>B>>C>>T){
    if(N==0&&M==0&&A==0&&B==0&&C==0&&T==0)break;
    solve();
  }
  return 0;
}
```