#include <bits/stdc++.h>
using namespace std;
int MOD;
vector<vector<int>> matrix_multiplication(vector<vector<int>> A, vector<vector<int>> B){
  int N = A.size();
  vector<vector<int>> ans(N, vector<int>(N, 0));
  for (int i = 0; i < N; i++){
    for (int j = 0; j < N; j++){
      for (int k = 0; k < N; k++){
        ans[i][k] += A[i][j] * B[j][k];
        ans[i][k] %= MOD;
      }
    }
  }
  return ans;
}
vector<vector<int>> matrix_exponentiation(vector<vector<int>> A, int b){
  int N = A.size();
  vector<vector<int>> ans(N, vector<int>(N, 0));
  for (int i = 0; i < N; i++){
    ans[i][i] = 1;
  }
  while (b > 0){
    if (b % 2 == 1){
      ans = matrix_multiplication(ans, A);
    }
    A = matrix_multiplication(A, A);
    b /= 2;
  }
  return ans;
}
int main(){
  while (1){
    int N, M, A, B, C, T;
    cin >> N >> M >> A >> B >> C >> T;
    if (N == 0 && M == 0 && A == 0 && B == 0 && C == 0 && T == 0){
      break;
    }
    vector<int> S(N + 2, 0);
    for (int i = 1; i <= N; i++){
      cin >> S[i];
    }
    MOD = M;
    vector<vector<int>> mat(N + 2, vector<int>(N + 2, 0));
    for (int i = 1; i <= N; i++){
      mat[i][i - 1] = A;
      mat[i][i] = B;
      mat[i][i + 1] = C;
    }
    mat = matrix_exponentiation(mat, T);
    vector<int> ans(N + 2, 0);
    for (int i = 0; i < N + 2; i++){
      for (int j = 0; j < N + 2; j++){
        ans[i] += mat[i][j] * S[j];
        ans[i] %= MOD;
      }
    }
    for (int i = 1; i <= N; i++){
      cout << ans[i];
      if (i < N){
        cout << ' ';
      }
    }
    cout << endl;
  }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 <bits/stdc++.h>
using namespace std;
int MOD;
vector<vector<int>> matrix_multiplication(vector<vector<int>> A, vector<vector<int>> B){
  int N = A.size();
  vector<vector<int>> ans(N, vector<int>(N, 0));
  for (int i = 0; i < N; i++){
    for (int j = 0; j < N; j++){
      for (int k = 0; k < N; k++){
        ans[i][k] += A[i][j] * B[j][k];
        ans[i][k] %= MOD;
      }
    }
  }
  return ans;
}
vector<vector<int>> matrix_exponentiation(vector<vector<int>> A, int b){
  int N = A.size();
  vector<vector<int>> ans(N, vector<int>(N, 0));
  for (int i = 0; i < N; i++){
    ans[i][i] = 1;
  }
  while (b > 0){
    if (b % 2 == 1){
      ans = matrix_multiplication(ans, A);
    }
    A = matrix_multiplication(A, A);
    b /= 2;
  }
  return ans;
}
int main(){
  while (1){
    int N, M, A, B, C, T;
    cin >> N >> M >> A >> B >> C >> T;
    if (N == 0 && M == 0 && A == 0 && B == 0 && C == 0 && T == 0){
      break;
    }
    vector<int> S(N + 2, 0);
    for (int i = 1; i <= N; i++){
      cin >> S[i];
    }
    MOD = M;
    vector<vector<int>> mat(N + 2, vector<int>(N + 2, 0));
    for (int i = 1; i <= N; i++){
      mat[i][i - 1] = A;
      mat[i][i] = B;
      mat[i][i + 1] = C;
    }
    mat = matrix_exponentiation(mat, T);
    vector<int> ans(N + 2, 0);
    for (int i = 0; i < N + 2; i++){
      for (int j = 0; j < N + 2; j++){
        ans[i] += mat[i][j] * S[j];
        ans[i] %= MOD;
      }
    }
    for (int i = 1; i <= N; i++){
      cout << ans[i];
      if (i < N){
        cout << ' ';
      }
    }
    cout << endl;
  }
}

```