#include <cstdio>
#include <utility>
#include <algorithm>
#include <vector>
#include <cstdint>

using Matrix=std::vector<std::vector<intmax_t>>;

Matrix matpow(Matrix &Q, intmax_t T, intmax_t M) {
  Matrix zero=Matrix(Q.size());
  for (size_t i=0; i<zero.size(); ++i)
    zero[i].resize(Q.size());

  Matrix eye=zero;
  for (size_t i=0; i<eye.size(); ++i)
    eye[i][i] = 1;

  Matrix res=eye;
  Matrix pow=Q;

  while (T) {
    if (T & 1) {
      // res *= pow;
      Matrix tmp=zero;
      for (size_t i=0; i<Q.size(); ++i)
	for (size_t j=0; j<Q.size(); ++j)
	  for (size_t k=0; k<Q.size(); ++k) {
	    tmp[i][j] += res[i][k] * pow[k][j];
	    tmp[i][j] %= M;
	  }

      res = std::move(tmp);
    }

    {
      // pow *= pow;
      Matrix tmp=zero;
      for (size_t i=0; i<Q.size(); ++i)
	for (size_t j=0; j<Q.size(); ++j)
	  for (size_t k=0; k<Q.size(); ++k) {
	    tmp[i][j] += pow[i][k] * pow[k][j];
	    tmp[i][j] %= M;
	  }
    
      pow = std::move(tmp);
    }

    T >>= 1;
  }

  return res;
}

int testcase_ends() {
  intmax_t N, M, A, B, C, T;
  scanf("%jd %jd %jd %jd %jd %jd", &N, &M, &A, &B, &C, &T);

  if (N == 0)
    return 1;

  std::vector<intmax_t> S(N+2);
  for (int i=1; i<=N; ++i)
    scanf("%jd", &S[i]);

  std::vector<std::vector<intmax_t>> Q(N+2, std::vector<intmax_t>(N+2));
  for (int i=1; i<N+1; ++i) Q[i][i] = B;
  for (int i=1; i<N+1; ++i) {
    Q[i][i-1] = A;
    Q[i][i+1] = C;
  }

  Matrix R=matpow(Q, T, M);

  std::vector<intmax_t> res(N+2);
  for (int i=0; i<=N+1; ++i)
    for (int j=0; j<=N+1; ++j) {
      res[i] += R[i][j] * S[j];
      res[i] %= M;
    }

  for (int i=1; i<=N; ++i)
    printf("%jd%c", res[i], i<N? ' ':'\n');

  return 0;
}

int main() {
  while (!testcase_ends()) {}

  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 <cstdio>
#include <utility>
#include <algorithm>
#include <vector>
#include <cstdint>

using Matrix=std::vector<std::vector<intmax_t>>;

Matrix matpow(Matrix &Q, intmax_t T, intmax_t M) {
  Matrix zero=Matrix(Q.size());
  for (size_t i=0; i<zero.size(); ++i)
    zero[i].resize(Q.size());

  Matrix eye=zero;
  for (size_t i=0; i<eye.size(); ++i)
    eye[i][i] = 1;

  Matrix res=eye;
  Matrix pow=Q;

  while (T) {
    if (T & 1) {
      // res *= pow;
      Matrix tmp=zero;
      for (size_t i=0; i<Q.size(); ++i)
	for (size_t j=0; j<Q.size(); ++j)
	  for (size_t k=0; k<Q.size(); ++k) {
	    tmp[i][j] += res[i][k] * pow[k][j];
	    tmp[i][j] %= M;
	  }

      res = std::move(tmp);
    }

    {
      // pow *= pow;
      Matrix tmp=zero;
      for (size_t i=0; i<Q.size(); ++i)
	for (size_t j=0; j<Q.size(); ++j)
	  for (size_t k=0; k<Q.size(); ++k) {
	    tmp[i][j] += pow[i][k] * pow[k][j];
	    tmp[i][j] %= M;
	  }
    
      pow = std::move(tmp);
    }

    T >>= 1;
  }

  return res;
}

int testcase_ends() {
  intmax_t N, M, A, B, C, T;
  scanf("%jd %jd %jd %jd %jd %jd", &N, &M, &A, &B, &C, &T);

  if (N == 0)
    return 1;

  std::vector<intmax_t> S(N+2);
  for (int i=1; i<=N; ++i)
    scanf("%jd", &S[i]);

  std::vector<std::vector<intmax_t>> Q(N+2, std::vector<intmax_t>(N+2));
  for (int i=1; i<N+1; ++i) Q[i][i] = B;
  for (int i=1; i<N+1; ++i) {
    Q[i][i-1] = A;
    Q[i][i+1] = C;
  }

  Matrix R=matpow(Q, T, M);

  std::vector<intmax_t> res(N+2);
  for (int i=0; i<=N+1; ++i)
    for (int j=0; j<=N+1; ++j) {
      res[i] += R[i][j] * S[j];
      res[i] %= M;
    }

  for (int i=1; i<=N; ++i)
    printf("%jd%c", res[i], i<N? ' ':'\n');

  return 0;
}

int main() {
  while (!testcase_ends()) {}

  return 0;
}
```