#include <bits/stdc++.h>

using namespace std;

#define rep(i, n) for (int i = 0; i < int(n); ++i)

typedef vector<long long> V;
typedef vector<V> M;

long long MOD;

V mul(M m, V v) {
  V res(m.size(), 0);
  rep (i, m.size()) rep (j, v.size()) {
    res[i] += m[i][j] * v[j];
    res[i] %= MOD;
  }
  return res;
}

M mul(const M &a, const M &b) {
  M res(a.size(), V(b[0].size(), 0));
  rep (i, a.size()) rep (j, b[0].size()) rep (k, b.size()) {
    res[i][j] += a[i][k] * b[k][j];
    res[i][j] %= MOD;
  }
  return res;
}

M pow(const M &m, long long k) {
  if (k == 0) {
    M res(m.size(), V(m.size(), 0));
    rep (i, m.size()) res[i][i] = 1;
    return res;
  }
  M a = pow(m, k / 2);
  a = mul(a, a);
  if (k % 2 == 1) a = mul(a, m);
  return a;
}

int main() {
  while (true) {
    int n, a, b, c, t;
    cin >> n >> MOD >> a >> b >> c >> t;
    if (n == 0) break;
    M m(n, V(n, 0));
    rep (i, n - 1) m[i + 1][i] = a;
    rep (i, n) m[i][i] = b;
    rep (i, n - 1) m[i][i + 1] = c;
    m = pow(m, t);
    V v(n);
    rep (i, n) cin >> v[i];
    v = mul(m, v);
    rep (i, v.size()) cout << v[i] << (i == (int)v.size() - 1 ? "\n" : " ");
  }
}
### Prompt

Your challenge is to write 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 <bits/stdc++.h>

using namespace std;

#define rep(i, n) for (int i = 0; i < int(n); ++i)

typedef vector<long long> V;
typedef vector<V> M;

long long MOD;

V mul(M m, V v) {
  V res(m.size(), 0);
  rep (i, m.size()) rep (j, v.size()) {
    res[i] += m[i][j] * v[j];
    res[i] %= MOD;
  }
  return res;
}

M mul(const M &a, const M &b) {
  M res(a.size(), V(b[0].size(), 0));
  rep (i, a.size()) rep (j, b[0].size()) rep (k, b.size()) {
    res[i][j] += a[i][k] * b[k][j];
    res[i][j] %= MOD;
  }
  return res;
}

M pow(const M &m, long long k) {
  if (k == 0) {
    M res(m.size(), V(m.size(), 0));
    rep (i, m.size()) res[i][i] = 1;
    return res;
  }
  M a = pow(m, k / 2);
  a = mul(a, a);
  if (k % 2 == 1) a = mul(a, m);
  return a;
}

int main() {
  while (true) {
    int n, a, b, c, t;
    cin >> n >> MOD >> a >> b >> c >> t;
    if (n == 0) break;
    M m(n, V(n, 0));
    rep (i, n - 1) m[i + 1][i] = a;
    rep (i, n) m[i][i] = b;
    rep (i, n - 1) m[i][i + 1] = c;
    m = pow(m, t);
    V v(n);
    rep (i, n) cin >> v[i];
    v = mul(m, v);
    rep (i, v.size()) cout << v[i] << (i == (int)v.size() - 1 ? "\n" : " ");
  }
}
```