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

Solution (in CPP):

Solution:
#include<iostream>
#include<vector>

using namespace std;

typedef vector<vector<int> > MAT;

int M;

MAT make_mat(int x){
  return MAT(x,vector<int>(x));
}

MAT operator*(const MAT &a,const MAT &b){
  const int n=a.size();
  MAT r=make_mat(n);
  for(int i=0;i<n;i++){
    for(int j=0;j<n;j++){
      for(int k=0;k<n;k++){
	r[i][j]=(r[i][j]+a[i][k]*b[k][j])%M;
      }
    }
  }
  return r;
}

MAT pow(MAT m,int x){
  const int n=m.size();
  MAT r=make_mat(n);
  for(int i=0;i<n;i++){
    r[i][i]=1;
  }
  for(;x;x>>=1,m=m*m){
    if(x%2){
      r=r*m;
    }
  }
  return r;
}

int main(){
  for(int n,a,b,c,t;cin>>n>>M>>a>>b>>c>>t,n;){
    MAT m=make_mat(n);
    for(int i=0;i<n;i++){
      for(int j=0;j<n;j++){
	m[i][j]=(i-1==j)?a:
	  (i==j)?b:
	  (i+1==j)?c:0;
      }
    }
    m=pow(m,t);
    vector<int> s(n);
    for(int i=0;i<n;i++){
      cin>>s[i];
    }
    for(int i=0;i<n;i++){
      int ans=0;
      for(int j=0;j<n;j++){
	ans=(ans+m[i][j]*s[j])%M;
      }
      cout<<ans<<((i==n-1)?'\n':' ');
    }
  }
}