#include <iostream>
#include <algorithm>
using namespace std;
#define N 60
int n,m,a,b,c,t;
int ans[N],s[N];
void mult(int C[N][N], int A[N][N], int B[N][N]){
  int tmp[N][N];
  for(int i=0;i<n;++i)
    for(int j=0;j<n;++j){
      tmp[i][j]=0;
      for(int k=0;k<n;++k)
	tmp[i][j]=(tmp[i][j]+A[i][k]*B[k][j])%m;
    }
  for(int i=0;i<n;++i)
    for(int j=0;j<n;++j) C[i][j]=tmp[i][j];
}

void solve(){
  int A[N][N],M[N][N];
  for(int i=0;i<n;++i)
    for(int j=0;j<n;++j) A[i][j] = 0, M[i][j]=(i==j);
  for(int i=0;i<n;++i) A[i][i] = b;
  for(int i=1;i<n;++i){
    A[i][i-1]=a;
    A[i-1][i]=c;
  }

  while(t){
    if(t&1)
      mult(M,M,A);
    mult(A,A,A);
    t/=2;
  }

  for(int i=0;i<n;++i){
    ans[i]=0;
    for(int j=0;j<n;++j)
      ans[i]=(ans[i]+M[i][j]*s[j])%m;
  }
}

int main(){
  while(cin>>n>>m>>a>>b>>c>>t,n||m||a||b||c||t){
    for(int i=0;i<n;++i) cin >> s[i];
    solve();
    for(int i=0;i<n-1;++i) cout << ans[i] << " ";
    cout << ans[n-1] << endl;
  }
  return 0;
}
### Prompt

In CPP, your task is to solve 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 <algorithm>
using namespace std;
#define N 60
int n,m,a,b,c,t;
int ans[N],s[N];
void mult(int C[N][N], int A[N][N], int B[N][N]){
  int tmp[N][N];
  for(int i=0;i<n;++i)
    for(int j=0;j<n;++j){
      tmp[i][j]=0;
      for(int k=0;k<n;++k)
	tmp[i][j]=(tmp[i][j]+A[i][k]*B[k][j])%m;
    }
  for(int i=0;i<n;++i)
    for(int j=0;j<n;++j) C[i][j]=tmp[i][j];
}

void solve(){
  int A[N][N],M[N][N];
  for(int i=0;i<n;++i)
    for(int j=0;j<n;++j) A[i][j] = 0, M[i][j]=(i==j);
  for(int i=0;i<n;++i) A[i][i] = b;
  for(int i=1;i<n;++i){
    A[i][i-1]=a;
    A[i-1][i]=c;
  }

  while(t){
    if(t&1)
      mult(M,M,A);
    mult(A,A,A);
    t/=2;
  }

  for(int i=0;i<n;++i){
    ans[i]=0;
    for(int j=0;j<n;++j)
      ans[i]=(ans[i]+M[i][j]*s[j])%m;
  }
}

int main(){
  while(cin>>n>>m>>a>>b>>c>>t,n||m||a||b||c||t){
    for(int i=0;i<n;++i) cin >> s[i];
    solve();
    for(int i=0;i<n-1;++i) cout << ans[i] << " ";
    cout << ans[n-1] << endl;
  }
  return 0;
}
```