#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<string>
#include<vector>
#include<map>
#include<list>
#include<queue>
#include<deque>
#include<algorithm>
#include<numeric>
#include<utility>
#include<complex>
#include<functional>
 
using namespace std;

/* constant */

#define MAX_N	(50)

/* typedef */

typedef int mat_t[MAX_N][MAX_N];
typedef int vec_t[MAX_N];

/* subroutines */

void mulmod(int n, mat_t m, mat_t m1, int mod) {
  mat_t m0;
  memcpy(m0, m, sizeof(mat_t));

  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      int mij = 0;
      for (int k = 0; k < n; k++)
        mij += m0[i][k] * m1[k][j];
      m[i][j] = mij % mod;
    }
}

void powmod(int n, mat_t m, mat_t m0, int t, int mod) {
  memset(m, 0, sizeof(mat_t));
  for (int i = 0; i < n; i++) m[i][i] = 1;

  while (t > 0) {
    if (t & 1) mulmod(n, m, m0, mod);

    mat_t m1;
    memcpy(m1, m0, sizeof(mat_t));
    mulmod(n, m0, m1, mod);

    t >>= 1;
  }
}

/* main */

int main() {
  for (;;) {
    int n, m, a, b, c, t;
    cin >> n >> m >> a >> b >> c >> t;
    if ((n | m | a | b | c | t) == 0) break;

    vec_t v;
    for (int i = 0; i < n; i++) cin >> v[i];

    mat_t mat0;
    memset(mat0, 0, sizeof(mat_t));
  
    for (int i = 0; i < n; i++) {
      if (i > 0) mat0[i][i - 1] = a;
      mat0[i][i] = b;
      if (i < n - 1) mat0[i][i + 1] = c;
    }

    mat_t mat;
    powmod(n, mat, mat0, t, m);

    vec_t v0;
    for (int i = 0; i < n; i++) {
      int vi = 0;
      for (int j = 0; j < n; j++)
	vi += mat[i][j] * v[j];
      v0[i] = vi % m;
    }

    for (int i = 0; i < n; i++) {
      if (i > 0) cout << ' ';
      cout << v0[i];
    }
    cout << endl;
  }

  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<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<string>
#include<vector>
#include<map>
#include<list>
#include<queue>
#include<deque>
#include<algorithm>
#include<numeric>
#include<utility>
#include<complex>
#include<functional>
 
using namespace std;

/* constant */

#define MAX_N	(50)

/* typedef */

typedef int mat_t[MAX_N][MAX_N];
typedef int vec_t[MAX_N];

/* subroutines */

void mulmod(int n, mat_t m, mat_t m1, int mod) {
  mat_t m0;
  memcpy(m0, m, sizeof(mat_t));

  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      int mij = 0;
      for (int k = 0; k < n; k++)
        mij += m0[i][k] * m1[k][j];
      m[i][j] = mij % mod;
    }
}

void powmod(int n, mat_t m, mat_t m0, int t, int mod) {
  memset(m, 0, sizeof(mat_t));
  for (int i = 0; i < n; i++) m[i][i] = 1;

  while (t > 0) {
    if (t & 1) mulmod(n, m, m0, mod);

    mat_t m1;
    memcpy(m1, m0, sizeof(mat_t));
    mulmod(n, m0, m1, mod);

    t >>= 1;
  }
}

/* main */

int main() {
  for (;;) {
    int n, m, a, b, c, t;
    cin >> n >> m >> a >> b >> c >> t;
    if ((n | m | a | b | c | t) == 0) break;

    vec_t v;
    for (int i = 0; i < n; i++) cin >> v[i];

    mat_t mat0;
    memset(mat0, 0, sizeof(mat_t));
  
    for (int i = 0; i < n; i++) {
      if (i > 0) mat0[i][i - 1] = a;
      mat0[i][i] = b;
      if (i < n - 1) mat0[i][i + 1] = c;
    }

    mat_t mat;
    powmod(n, mat, mat0, t, m);

    vec_t v0;
    for (int i = 0; i < n; i++) {
      int vi = 0;
      for (int j = 0; j < n; j++)
	vi += mat[i][j] * v[j];
      v0[i] = vi % m;
    }

    for (int i = 0; i < n; i++) {
      if (i > 0) cout << ' ';
      cout << v0[i];
    }
    cout << endl;
  }

  return 0;
}
  
```