#include <iostream>
#include <vector>
#define N_MAX 50
using namespace std;
typedef long long ll;
typedef vector<ll> vec;
typedef vector<vec> mat;
ll N,M,A,B,C,T,d[N_MAX];

mat cal(mat mA,mat mB){
  mat res=mat(N,vec(N,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]+=mA[i][k]*mB[k][j]%M;
  return res;
}

void mod_power(mat &mA){
  mat res=mat(N,vec(N,0));
  for(int i=0;i<N;i++)
    for(int j=0;j<N;j++) res[i][j]=(i==j);
  for(int i=0;i<=30;i++){
    if(T&(1LL<<i)) res=cal(res,mA);
    mA=cal(mA,mA);
  }
  for(int i=0;i<N;i++)
    for(int j=0;j<N;j++) mA[i][j]=res[i][j];
}

int main(){
  while(1){
    cin>>N>>M>>A>>B>>C>>T;
    if(!N&&!M&&!A&&!B&&!C&&!T) break;
    mat mA=mat(N,vec(N,0));
    for(int i=0;i<N;i++) cin>>d[i];
    for(int i=0;i<N;i++)
      for(int j=0;j<N;j++){
	if(i==j) mA[i][j]=B;
	if(i+1==j) mA[i][j]=C;
	if(i==j+1) mA[i][j]=A;
      }
    mod_power(mA);
    mat mB=mat(N,vec(N,0));
    for(int i=0;i<N;i++) mB[i][0]=d[i];
    mA=cal(mA,mB);
    for(int i=0;i<N;i++){
      cout<<mA[i][0]%M;
      if(i!=N-1) cout<<' ';
    }
    cout<<endl;
  }
  return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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>
#define N_MAX 50
using namespace std;
typedef long long ll;
typedef vector<ll> vec;
typedef vector<vec> mat;
ll N,M,A,B,C,T,d[N_MAX];

mat cal(mat mA,mat mB){
  mat res=mat(N,vec(N,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]+=mA[i][k]*mB[k][j]%M;
  return res;
}

void mod_power(mat &mA){
  mat res=mat(N,vec(N,0));
  for(int i=0;i<N;i++)
    for(int j=0;j<N;j++) res[i][j]=(i==j);
  for(int i=0;i<=30;i++){
    if(T&(1LL<<i)) res=cal(res,mA);
    mA=cal(mA,mA);
  }
  for(int i=0;i<N;i++)
    for(int j=0;j<N;j++) mA[i][j]=res[i][j];
}

int main(){
  while(1){
    cin>>N>>M>>A>>B>>C>>T;
    if(!N&&!M&&!A&&!B&&!C&&!T) break;
    mat mA=mat(N,vec(N,0));
    for(int i=0;i<N;i++) cin>>d[i];
    for(int i=0;i<N;i++)
      for(int j=0;j<N;j++){
	if(i==j) mA[i][j]=B;
	if(i+1==j) mA[i][j]=C;
	if(i==j+1) mA[i][j]=A;
      }
    mod_power(mA);
    mat mB=mat(N,vec(N,0));
    for(int i=0;i<N;i++) mB[i][0]=d[i];
    mA=cal(mA,mB);
    for(int i=0;i<N;i++){
      cout<<mA[i][0]%M;
      if(i!=N-1) cout<<' ';
    }
    cout<<endl;
  }
  return 0;
}
```