#include <bits/stdc++.h>

using namespace std;

typedef vector<int> row;
typedef vector<row> matrix;

matrix multi(const matrix& a, const matrix& b, const int mod)
{
  matrix c(a.size(), row(b[0].size(), 0));
  for(int i = 0; i < a.size(); i++) {
    for(int j = 0; j < b.size(); j++) {
      for(int k = 0; k < b[0].size(); k++) {
	c[i][k] = (c[i][k] + a[i][j]*b[j][k]) % mod;
      }
    }
  }
  return c;
}

matrix pow(matrix a, int n, const int mod)
{
  matrix b(a.size(), row(a.size(), 0));
  for(int i = 0; i < a.size(); i++) b[i][i] = 1;
  while(n > 0) {
    if(n & 1) b = multi(b, a, mod);
    a = multi(a, a, mod);
    n >>= 1;
  }
  return b;
}

int main()
{
  int N, M, A, B, C, T;
  while(cin >> N >> M >> A >> B >> C >> T, N) {
    matrix a(N, row(N, 0)), b(N, row(1, 0));
    for(int i = 0; i < N; i++) {
      if(i > 0) a[i][i-1] = A;
      a[i][i] = B;
      if(i+1 < N) a[i][i+1] = C;
    }
    for(int i = 0; i < N; i++) cin >> b[i][0];
    a = multi(pow(a, T, M), b, M);
    for(int i = 0; i < N; i++) cout << a[i][0] << (i == N-1 ? '\n' : ' ');
  }
  return 0;
}
### Prompt

Create a solution in CPP for 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 <bits/stdc++.h>

using namespace std;

typedef vector<int> row;
typedef vector<row> matrix;

matrix multi(const matrix& a, const matrix& b, const int mod)
{
  matrix c(a.size(), row(b[0].size(), 0));
  for(int i = 0; i < a.size(); i++) {
    for(int j = 0; j < b.size(); j++) {
      for(int k = 0; k < b[0].size(); k++) {
	c[i][k] = (c[i][k] + a[i][j]*b[j][k]) % mod;
      }
    }
  }
  return c;
}

matrix pow(matrix a, int n, const int mod)
{
  matrix b(a.size(), row(a.size(), 0));
  for(int i = 0; i < a.size(); i++) b[i][i] = 1;
  while(n > 0) {
    if(n & 1) b = multi(b, a, mod);
    a = multi(a, a, mod);
    n >>= 1;
  }
  return b;
}

int main()
{
  int N, M, A, B, C, T;
  while(cin >> N >> M >> A >> B >> C >> T, N) {
    matrix a(N, row(N, 0)), b(N, row(1, 0));
    for(int i = 0; i < N; i++) {
      if(i > 0) a[i][i-1] = A;
      a[i][i] = B;
      if(i+1 < N) a[i][i+1] = C;
    }
    for(int i = 0; i < N; i++) cin >> b[i][0];
    a = multi(pow(a, T, M), b, M);
    for(int i = 0; i < N; i++) cout << a[i][0] << (i == N-1 ? '\n' : ' ');
  }
  return 0;
}
```