#include <bits/stdc++.h>
typedef long long LL;
#define SORT(c) sort((c).begin(),(c).end())

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)

using namespace std;
LL n,m,a,b,c,t;
vector<vector<LL> > matprod(vector<vector<LL> > mata,vector<vector<LL> > matb)
{
  vector<vector<LL> > ans;
  ans.resize(n);
  REP(i,n) ans[i].assign(n,0);
  REP(i,n) REP(j,n) REP(k,n)
    ans[i][k]=(ans[i][k]+(mata[i][j]*matb[j][k])%m)%m;
  return ans;
}
vector<vector<LL> > power(vector<vector<LL> > mat, int x)
{
  if(x==1)return mat;
  else if(x%2==1) return matprod(power(matprod(mat,mat),x/2),mat);
  else return power(matprod(mat,mat),x/2);
}
int main(void)
{
  for(;;){
    cin >> n >> m >> a >> b >> c >> t;
    if(!n&&!m) return 0;
    vector<LL> s;
    s.resize(n);
    REP(i,n) cin >> s[i];
    vector<vector<LL> > mat;
    mat.resize(n);
    REP(i,n) mat[i].resize(n);
    REP(i,n) REP(j,n){
      if(i==j-1) mat[i][j]=a;
      else if(i==j) mat[i][j]=b;
      else if(i==j+1) mat[i][j]=c;
      else mat[i][j]=0;
    }
    vector<LL> ans;
    if(t==0) ans=vector<LL>(s);
    else{
      ans.assign(n,0);
      vector<vector<LL> > mat1=power(mat,t);
      REP(i,n) REP(j,n) ans[j]=(ans[j]+(mat1[i][j]*s[i])%m)%m;
    }
    REP(i,n){
      cout << ans[i];
      if(i==n-1) cout << endl;
      else cout << ' ';
    }
  }
}
### Prompt

Your task is to create a cpp solution 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>
typedef long long LL;
#define SORT(c) sort((c).begin(),(c).end())

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)

using namespace std;
LL n,m,a,b,c,t;
vector<vector<LL> > matprod(vector<vector<LL> > mata,vector<vector<LL> > matb)
{
  vector<vector<LL> > ans;
  ans.resize(n);
  REP(i,n) ans[i].assign(n,0);
  REP(i,n) REP(j,n) REP(k,n)
    ans[i][k]=(ans[i][k]+(mata[i][j]*matb[j][k])%m)%m;
  return ans;
}
vector<vector<LL> > power(vector<vector<LL> > mat, int x)
{
  if(x==1)return mat;
  else if(x%2==1) return matprod(power(matprod(mat,mat),x/2),mat);
  else return power(matprod(mat,mat),x/2);
}
int main(void)
{
  for(;;){
    cin >> n >> m >> a >> b >> c >> t;
    if(!n&&!m) return 0;
    vector<LL> s;
    s.resize(n);
    REP(i,n) cin >> s[i];
    vector<vector<LL> > mat;
    mat.resize(n);
    REP(i,n) mat[i].resize(n);
    REP(i,n) REP(j,n){
      if(i==j-1) mat[i][j]=a;
      else if(i==j) mat[i][j]=b;
      else if(i==j+1) mat[i][j]=c;
      else mat[i][j]=0;
    }
    vector<LL> ans;
    if(t==0) ans=vector<LL>(s);
    else{
      ans.assign(n,0);
      vector<vector<LL> > mat1=power(mat,t);
      REP(i,n) REP(j,n) ans[j]=(ans[j]+(mat1[i][j]*s[i])%m)%m;
    }
    REP(i,n){
      cout << ans[i];
      if(i==n-1) cout << endl;
      else cout << ' ';
    }
  }
}
```