#include <vector>
#include <algorithm>
#include <iostream>
#include <cstdio>

using namespace std;

typedef long long ll;

typedef vector<vector<ll>> Mat;

Mat createMat(int N, int A, int B, int C) {
  Mat mat;

  for (int i=0; i<N; i++) {
    mat.push_back(vector<ll>(N));
    if (0 <= i-1) mat[i][i-1] = A;
    mat[i][i] = B;
    if (i+1 < N) mat[i][i+1] = C;
  }

  return mat;
}

Mat mul(Mat &A, Mat &B, int M) {
  int N = A.size();
  Mat res(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++) {
        res[i][j] += A[i][k] * B[k][j];
        res[i][j] %= M;
      }
    }
  }

  return res;
}

Mat power(Mat &A, int n, int M) {
  Mat res, p = A;

  int N = A.size();
  for (int i=0; i<N; i++) {
    res.push_back(vector<ll>(N));
    res[i][i] = 1;
  }


  while(n) {
    if (n % 2) {
      res = mul(res, p, M);
    }
    p = mul(p, p, M);
    n /= 2;
  }

  return res;
}

int main() {
  int S[100];

  while(1) {
    int N, M, A, B, C, T;

    scanf("%d%d%d%d%d%d", &N, &M, &A, &B, &C, &T);

    if (N == 0) break;

    Mat mat = createMat(N, A, B, C);

    Mat res = power(mat, T, M);

    for (int i=0; i<N; i++) {
      scanf("%d", S+i);
    }

    for (int i=0; i<N; i++) {
      ll ans = 0;

      for (int j=0; j<N; j++) {
        ans += res[i][j] * S[j] % M;
      }

      printf("%lld%c", ans % M, " \n"[i == N-1]);
    }
  }
}


### Prompt

Develop a solution in Cpp 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 <vector>
#include <algorithm>
#include <iostream>
#include <cstdio>

using namespace std;

typedef long long ll;

typedef vector<vector<ll>> Mat;

Mat createMat(int N, int A, int B, int C) {
  Mat mat;

  for (int i=0; i<N; i++) {
    mat.push_back(vector<ll>(N));
    if (0 <= i-1) mat[i][i-1] = A;
    mat[i][i] = B;
    if (i+1 < N) mat[i][i+1] = C;
  }

  return mat;
}

Mat mul(Mat &A, Mat &B, int M) {
  int N = A.size();
  Mat res(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++) {
        res[i][j] += A[i][k] * B[k][j];
        res[i][j] %= M;
      }
    }
  }

  return res;
}

Mat power(Mat &A, int n, int M) {
  Mat res, p = A;

  int N = A.size();
  for (int i=0; i<N; i++) {
    res.push_back(vector<ll>(N));
    res[i][i] = 1;
  }


  while(n) {
    if (n % 2) {
      res = mul(res, p, M);
    }
    p = mul(p, p, M);
    n /= 2;
  }

  return res;
}

int main() {
  int S[100];

  while(1) {
    int N, M, A, B, C, T;

    scanf("%d%d%d%d%d%d", &N, &M, &A, &B, &C, &T);

    if (N == 0) break;

    Mat mat = createMat(N, A, B, C);

    Mat res = power(mat, T, M);

    for (int i=0; i<N; i++) {
      scanf("%d", S+i);
    }

    for (int i=0; i<N; i++) {
      ll ans = 0;

      for (int j=0; j<N; j++) {
        ans += res[i][j] * S[j] % M;
      }

      printf("%lld%c", ans % M, " \n"[i == N-1]);
    }
  }
}


```