#include <iostream>
#include <vector>
typedef long long ll;
using namespace std;
int N,M;
vector<vector <ll> > gyouretu1(vector<vector <ll> > a,vector<vector<ll> > b){
  vector< vector<ll> > ans(N,vector<ll>(N));
  for(int i = 0;i < N;i++){
    for(int j = 0;j < N;j++){
      for(int k = 0;k < N;k++){
	ans[i][j] = (ans[i][j] + a[k][j]*b[i][k]);
      }
      ans[i][j] = ans[i][j] % M;
    }
  }
  return ans;
}

int main(){
  while(true){
    int A,B,C,T;
    cin >> N;
    cin >> M;
    cin >> A;
    cin >> B;
    cin >> C;
    cin >> T;
    vector<ll> s  = vector<ll>(N);
    if(N == 0 && M == 0 && A == 0 && B == 0 && C == 0 && T == 0)break;
    for(int i = 0;i < N;i++){
      cin >> s[i];
    }
    vector< vector<ll> > x(N,vector<ll>(N));
    int a,b,c;
    a = -1;b = 0;c = 1; 
    
    for(int i = 0;i < N;i++){
      for(int j = 0;j < N;j++){
	if(j == a){
	  x[i][j] = A;  
	}else if(j == b){
	  x[i][j] = B;
	}else if(j == c){
	  x[i][j] = C;
	}else x[i][j] = 0;
      }
      a++;
      b++;
      c++;
    }
    vector< vector<ll> > ans(N,vector<ll>(N));
    for(int i = 0;i < N;i++){
      for(int j = 0;j < N;j++){
	if(j == i)ans[i][j] = 1;
	else ans[i][j] = 0;
      }
    }
    
    while(T > 0){
      if(T & 1){
	ans = gyouretu1(ans,x);
      }
      x = gyouretu1(x,x);
      T = T >> 1;
    }
    ll anser = 0;
    //cout << s[N-2] << "aaaaaaaa" << endl;
    
    for(int i = 0;i < N;i++){
      for(int j = 0;j < N;j++){
	anser = anser + ans[i][j]*s[j];
      }
      if(i == 0)
	cout << anser%M;
      else cout << " " << anser%M;
      anser = 0;
    }
    cout << endl;
    
  }
    
    
    

  
  return 0;
}
### Prompt

Please formulate 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>
typedef long long ll;
using namespace std;
int N,M;
vector<vector <ll> > gyouretu1(vector<vector <ll> > a,vector<vector<ll> > b){
  vector< vector<ll> > ans(N,vector<ll>(N));
  for(int i = 0;i < N;i++){
    for(int j = 0;j < N;j++){
      for(int k = 0;k < N;k++){
	ans[i][j] = (ans[i][j] + a[k][j]*b[i][k]);
      }
      ans[i][j] = ans[i][j] % M;
    }
  }
  return ans;
}

int main(){
  while(true){
    int A,B,C,T;
    cin >> N;
    cin >> M;
    cin >> A;
    cin >> B;
    cin >> C;
    cin >> T;
    vector<ll> s  = vector<ll>(N);
    if(N == 0 && M == 0 && A == 0 && B == 0 && C == 0 && T == 0)break;
    for(int i = 0;i < N;i++){
      cin >> s[i];
    }
    vector< vector<ll> > x(N,vector<ll>(N));
    int a,b,c;
    a = -1;b = 0;c = 1; 
    
    for(int i = 0;i < N;i++){
      for(int j = 0;j < N;j++){
	if(j == a){
	  x[i][j] = A;  
	}else if(j == b){
	  x[i][j] = B;
	}else if(j == c){
	  x[i][j] = C;
	}else x[i][j] = 0;
      }
      a++;
      b++;
      c++;
    }
    vector< vector<ll> > ans(N,vector<ll>(N));
    for(int i = 0;i < N;i++){
      for(int j = 0;j < N;j++){
	if(j == i)ans[i][j] = 1;
	else ans[i][j] = 0;
      }
    }
    
    while(T > 0){
      if(T & 1){
	ans = gyouretu1(ans,x);
      }
      x = gyouretu1(x,x);
      T = T >> 1;
    }
    ll anser = 0;
    //cout << s[N-2] << "aaaaaaaa" << endl;
    
    for(int i = 0;i < N;i++){
      for(int j = 0;j < N;j++){
	anser = anser + ans[i][j]*s[j];
      }
      if(i == 0)
	cout << anser%M;
      else cout << " " << anser%M;
      anser = 0;
    }
    cout << endl;
    
  }
    
    
    

  
  return 0;
}
```