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
#include <iostream>
#include <vector>
using namespace std;
typedef long long ll;
typedef vector<ll> vec;
typedef vector <vec> mat;
ll n,m,a,b,c,T;

mat mul(mat A,mat B){
  mat res;
  res.resize(n);
  for(int i=0;i<n;i++)
    for(int j=0;j<n;j++) res[i].push_back(0);

  for(int i=0;i<n;i++)
    for(int j=0;j<n;j++)
      for(int k=0;k<n;k++){
	res[i][j]+=A[i][k]*B[k][j];
	res[i][j]%=m;
      }
  return res;
}

mat mod_pow(mat x,ll y){
  mat res;
  res.resize(n);
  for(int i=0;i<n;i++)for(int j=0;j<n;j++)res[i].push_back(i==j);
  
  while(y){
    if(y%2) res=mul(res,x);
    x=mul(x,x);
    y/=2;
  }
  return res; 
}
int main(){ 
  while(1){
    cin>>n>>m>>a>>b>>c>>T;
    if(!n&&!m&&!a&&!b&&!c&&!T)break;

    mat s;
    s.resize(n);
    for(int i=0;i<n;i++) s[i].resize(n);
    for(int i=0;i<n;i++) cin>>s[i][0];
    
    mat x;
    x.resize(n);
    for(int i=0;i<n;i++) x[i].resize(n);

    for(int i=0;i<n;i++)
      for(int j=0;j<n;j++)
	if(i==j) x[i][j]=b;
	else if(i+1==j&&i+1<n) x[i][j]=c;
	else if(i==j+1&&j+1<n) x[i][j]=a;
	else x[i][j]=0;
    
    x=mod_pow(x,T);
    mat ans=mul(x,s);
    for(int i=0;i<n-1;i++)cout << ans[i][0]<<" ";
    cout << ans[n-1][0]<<endl;
  }
  return 0;
}