#include <iostream>
#include <cstdio>
#include <vector>
#include <list>
#include <cmath>
#include <fstream>
#include <algorithm>
#include <string>
#include <queue>
#include <set>
#include <map>
#include <complex>
#include <iterator>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <stack>
#include <climits>
#include <deque>
#include <bitset>

using namespace std;


typedef long long ll;
typedef pair<int,int> pii;
const double EPS=1e-8;
bool EQ(double a,double b){
  return abs(a-b)<EPS;
}

typedef vector<ll> vec;
typedef vector<vec> mat;

void fastStream(){
  cin.tie(0);
  std::ios_base::sync_with_stdio(0);
}

ll mod;
mat mulm(mat &A,mat &B){
  mat C(A.size(),vec(B[0].size()));
  for(int i=0;i<(int)A.size();i++){
    for(int k=0;k<(int)B.size();k++){
      for(int j=0;j<(int)B[0].size();j++){
        C[i][j]=(C[i][j]+A[i][k]*B[k][j])%mod;
      }
    }
  }
  return C;
}

mat pow(mat A,ll n){
  mat B(A.size(),vec(A.size()));
  for(int i=0;i<(int)A.size();i++){
    B[i][i]=1;
  }
  while(n>0){
    if(n&1)B=mulm(B,A);
    A=mulm(A,A);
    n>>=1;
  }
  return B;
}

ll N,M,A,B,C,T;
ll SS[1001];
int main(){
  fastStream();
  while(cin>>N>>M>>A>>B>>C>>T&&(N|M|A|B|C)){
    mod=M;
    for(int i=0;i<N;i++)cin>>SS[i];
    mat AA(N,vec(N,0));
    AA[0][0]=B;
    if(N>1)AA[0][1]=C;
    for(int i=1;i<N-1;i++){
      AA[i][i-1]=A;
      AA[i][i]=B;
      AA[i][i+1]=C;
    }
    if(N-2>=0)
      AA[N-1][N-2]=A;
    AA[N-1][N-1]=B;
    vec v(N,0);
    for(int i=0;i<N;i++)v[i]=SS[i];
    mat res=pow(AA,T);
    vector<ll> ans(N,0);

    for(int i=0;i<N;i++){
      for(int j=0;j<N;j++){
        ans[i]+=(res[i][j]*v[j])%mod;
        ans[i]%=mod;
      }
    }
    for(int i=0;i<N;i++){
      cout<<ans[i];
      if(i==N-1)cout<<endl;
      else cout<<' ';
    }
    
  }
  
  
  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 <iostream>
#include <cstdio>
#include <vector>
#include <list>
#include <cmath>
#include <fstream>
#include <algorithm>
#include <string>
#include <queue>
#include <set>
#include <map>
#include <complex>
#include <iterator>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <stack>
#include <climits>
#include <deque>
#include <bitset>

using namespace std;


typedef long long ll;
typedef pair<int,int> pii;
const double EPS=1e-8;
bool EQ(double a,double b){
  return abs(a-b)<EPS;
}

typedef vector<ll> vec;
typedef vector<vec> mat;

void fastStream(){
  cin.tie(0);
  std::ios_base::sync_with_stdio(0);
}

ll mod;
mat mulm(mat &A,mat &B){
  mat C(A.size(),vec(B[0].size()));
  for(int i=0;i<(int)A.size();i++){
    for(int k=0;k<(int)B.size();k++){
      for(int j=0;j<(int)B[0].size();j++){
        C[i][j]=(C[i][j]+A[i][k]*B[k][j])%mod;
      }
    }
  }
  return C;
}

mat pow(mat A,ll n){
  mat B(A.size(),vec(A.size()));
  for(int i=0;i<(int)A.size();i++){
    B[i][i]=1;
  }
  while(n>0){
    if(n&1)B=mulm(B,A);
    A=mulm(A,A);
    n>>=1;
  }
  return B;
}

ll N,M,A,B,C,T;
ll SS[1001];
int main(){
  fastStream();
  while(cin>>N>>M>>A>>B>>C>>T&&(N|M|A|B|C)){
    mod=M;
    for(int i=0;i<N;i++)cin>>SS[i];
    mat AA(N,vec(N,0));
    AA[0][0]=B;
    if(N>1)AA[0][1]=C;
    for(int i=1;i<N-1;i++){
      AA[i][i-1]=A;
      AA[i][i]=B;
      AA[i][i+1]=C;
    }
    if(N-2>=0)
      AA[N-1][N-2]=A;
    AA[N-1][N-1]=B;
    vec v(N,0);
    for(int i=0;i<N;i++)v[i]=SS[i];
    mat res=pow(AA,T);
    vector<ll> ans(N,0);

    for(int i=0;i<N;i++){
      for(int j=0;j<N;j++){
        ans[i]+=(res[i][j]*v[j])%mod;
        ans[i]%=mod;
      }
    }
    for(int i=0;i<N;i++){
      cout<<ans[i];
      if(i==N-1)cout<<endl;
      else cout<<' ';
    }
    
  }
  
  
  return 0;
}
```