#include<iostream>
#include<vector>
using namespace std;
int n,m,a,b,c,t;
struct ary{
  int x[50][50]; 
  ary(){}
  ary(const ary &p){
    for(int i=0;i<50;i++){
      for(int j=0;j<50;j++){
	x[i][j]=p.x[i][j];
      }
    }
  }
};

ary mp,st;
ary multi(ary bmp1,ary bmp2){
  ary nmp;
  for(int i=0;i<n;i++){
    for(int j=0;j<n;j++){
      nmp.x[i][j]=0;
      for(int k=0;k<n;k++){
	nmp.x[i][j]+=(bmp1.x[i][k]*bmp2.x[k][j])%m;
      }
      nmp.x[i][j]%=m;
    }
  }
  return nmp;
}

ary rec(ary bmp,int ti){
  ary nmp(bmp);
  ti--;
  while(ti>0){
    if(ti&1) nmp=multi(nmp,bmp);
    bmp=multi(bmp,bmp);
    ti>>=1;
  }
  return nmp;
}
int main(){
  while(cin>>n>>m>>a>>b>>c>>t,n||m||a||b||c||t){
    int s[50]={},o[50]={};
    int i,j,k,p,q;
    for(i=0;i<n;i++){
      cin >> s[i]; 
    }
    fill(mp.x[0],mp.x[50],0);
    fill(st.x[0],st.x[50],0);
    mp.x[0][0]=b; mp.x[0][1]=c;
    for(i=1;i<n-1;i++){
      mp.x[i][i-1]=a;
      mp.x[i][i]=b;
      mp.x[i][i+1]=c;
    }
    mp.x[n-1][n-2]=a; mp.x[n-1][n-1]=b;
    mp=rec(mp,t);
    if(t==0){
      for(i=0;i<n;i++)
	cout << s[i] << (i==n-1?"\n":" ");
    }else{
      for(i=0;i<n;i++){
	o[i]=0;
	for(k=0;k<n;k++){
	  o[i]+=(mp.x[i][k]*s[k])%m;
	}
	o[i]%=m;
	cout << o[i] << (i==n-1?"\n":" ");
      }
    }
  }
  return 0;
}
### 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<iostream>
#include<vector>
using namespace std;
int n,m,a,b,c,t;
struct ary{
  int x[50][50]; 
  ary(){}
  ary(const ary &p){
    for(int i=0;i<50;i++){
      for(int j=0;j<50;j++){
	x[i][j]=p.x[i][j];
      }
    }
  }
};

ary mp,st;
ary multi(ary bmp1,ary bmp2){
  ary nmp;
  for(int i=0;i<n;i++){
    for(int j=0;j<n;j++){
      nmp.x[i][j]=0;
      for(int k=0;k<n;k++){
	nmp.x[i][j]+=(bmp1.x[i][k]*bmp2.x[k][j])%m;
      }
      nmp.x[i][j]%=m;
    }
  }
  return nmp;
}

ary rec(ary bmp,int ti){
  ary nmp(bmp);
  ti--;
  while(ti>0){
    if(ti&1) nmp=multi(nmp,bmp);
    bmp=multi(bmp,bmp);
    ti>>=1;
  }
  return nmp;
}
int main(){
  while(cin>>n>>m>>a>>b>>c>>t,n||m||a||b||c||t){
    int s[50]={},o[50]={};
    int i,j,k,p,q;
    for(i=0;i<n;i++){
      cin >> s[i]; 
    }
    fill(mp.x[0],mp.x[50],0);
    fill(st.x[0],st.x[50],0);
    mp.x[0][0]=b; mp.x[0][1]=c;
    for(i=1;i<n-1;i++){
      mp.x[i][i-1]=a;
      mp.x[i][i]=b;
      mp.x[i][i+1]=c;
    }
    mp.x[n-1][n-2]=a; mp.x[n-1][n-1]=b;
    mp=rec(mp,t);
    if(t==0){
      for(i=0;i<n;i++)
	cout << s[i] << (i==n-1?"\n":" ");
    }else{
      for(i=0;i<n;i++){
	o[i]=0;
	for(k=0;k<n;k++){
	  o[i]+=(mp.x[i][k]*s[k])%m;
	}
	o[i]%=m;
	cout << o[i] << (i==n-1?"\n":" ");
      }
    }
  }
  return 0;
}
```