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

#define rep(i, n) for(int i = 0; i < (n); i++)
#define repp(i, l, r) for(int i = (l); i < (r); i++)
#define per(i, n) for(int i = ((n)-1); i >= 0; i--)
#define perr(i, l, r) for(int i = ((r)-1); i >= (l); i--)
#define all(x) (x).begin(),(x).end()
#define MOD 1000000007
#define IINF 1000000000
#define LINF 1000000000000000000
#define SP <<" "<<
#define CYES cout<<"Yes"<<endl
#define CNO cout<<"No"<<endl
#define CFS cin.tie(0);ios::sync_with_stdio(false)

typedef long long LL;
typedef long double LD;

void mulmat(vector<vector<int>> &a,vector<vector<int>> &b, int m){
  int n = b.size();
  vector<vector<int>> ans(n,vector<int>(n,0));
  rep(i,n){
    rep(j,n){
      rep(k,n){
        ans[i][j]=(ans[i][j]+a[i][k]*b[k][j])%m;
      }
    }
  }
  b=ans;
}

void mulv(vector<vector<int>> &a, vector<int> &v, int m){
  int n = v.size();
  vector<int> ans(n,0);
  rep(i,n){
    rep(j,n){
      ans[i]=(ans[i]+a[i][j]*v[j])%m;
    }
  }
  v=ans;
}

void solve(vector<int> &cell, int a, int b, int c, int t, int m){
  int n = cell.size();
  vector<vector<int>> mat(n,vector<int>(n,0));
  rep(i,n){
    mat[i][i]=b;
    if(i>0) mat[i][i-1]=a;
    if(i<n-1) mat[i][i+1]=c;
  }
  while(t){
    if(t&1){
      mulv(mat,cell,m);
    }
    mulmat(mat,mat,m);
    t>>=1;
  }
}

int main(){
  while(1){
    int n,m,a,b,c,t;
    cin >> n >> m >> a >> b >> c >> t;
    if(n==0) return 0;
    vector<int> cell(n);
    rep(i,n) cin >> cell[i];
    solve(cell,a,b,c,t,m);
    rep(i,n){
      cout << cell[i];
      if(i<n-1) cout << " ";
      else 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 <bits/stdc++.h>
using namespace std;

#define rep(i, n) for(int i = 0; i < (n); i++)
#define repp(i, l, r) for(int i = (l); i < (r); i++)
#define per(i, n) for(int i = ((n)-1); i >= 0; i--)
#define perr(i, l, r) for(int i = ((r)-1); i >= (l); i--)
#define all(x) (x).begin(),(x).end()
#define MOD 1000000007
#define IINF 1000000000
#define LINF 1000000000000000000
#define SP <<" "<<
#define CYES cout<<"Yes"<<endl
#define CNO cout<<"No"<<endl
#define CFS cin.tie(0);ios::sync_with_stdio(false)

typedef long long LL;
typedef long double LD;

void mulmat(vector<vector<int>> &a,vector<vector<int>> &b, int m){
  int n = b.size();
  vector<vector<int>> ans(n,vector<int>(n,0));
  rep(i,n){
    rep(j,n){
      rep(k,n){
        ans[i][j]=(ans[i][j]+a[i][k]*b[k][j])%m;
      }
    }
  }
  b=ans;
}

void mulv(vector<vector<int>> &a, vector<int> &v, int m){
  int n = v.size();
  vector<int> ans(n,0);
  rep(i,n){
    rep(j,n){
      ans[i]=(ans[i]+a[i][j]*v[j])%m;
    }
  }
  v=ans;
}

void solve(vector<int> &cell, int a, int b, int c, int t, int m){
  int n = cell.size();
  vector<vector<int>> mat(n,vector<int>(n,0));
  rep(i,n){
    mat[i][i]=b;
    if(i>0) mat[i][i-1]=a;
    if(i<n-1) mat[i][i+1]=c;
  }
  while(t){
    if(t&1){
      mulv(mat,cell,m);
    }
    mulmat(mat,mat,m);
    t>>=1;
  }
}

int main(){
  while(1){
    int n,m,a,b,c,t;
    cin >> n >> m >> a >> b >> c >> t;
    if(n==0) return 0;
    vector<int> cell(n);
    rep(i,n) cin >> cell[i];
    solve(cell,a,b,c,t,m);
    rep(i,n){
      cout << cell[i];
      if(i<n-1) cout << " ";
      else cout << endl;
    }
  }
  return 0;
}


```