#include <bits/stdc++.h>

using namespace std;
using ll = long long;
using pii = pair<int, int>;

using vec = vector<int>;
using mat = vector<vec>;
 
vec mul(mat &A, vec &B, ll mod) {
  vec C(A.size());
  for (int i = 0; i < A.size(); i++) for (int j = 0; j < B.size(); j++) {
    (C[i] += A[i][j] * B[j]) %= mod;
  }
  return C;
}

mat mul(mat &A, mat &B, ll mod) {
  mat C(A.size(), vec(B[0].size()));
  for (int i = 0; i < A.size(); i++) for (int k = 0; k < B.size(); k++) for (int j = 0; j < B[0].size(); j++) {
    (C[i][j] += A[i][k] * B[k][j]) %= mod;
  }
  return C;
}
 
mat mat_pow(mat A, ll n, ll mod) {
  mat B(A.size(), vec(A.size()));
  for (int i = 0; i < A.size(); i++) {
    B[i][i] = 1;
  }
  while (n) {
    if (n & 1) B = mul(B, A, mod);
    A = mul(A, A, mod);
    n >>= 1;
  }
  return B;
}

int N, M, A, B, C, T;

int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(10);
  
  while (cin >> N >> M >> A >> B >> C >> T, N) {
    mat D(N, vec(N));
    for (int i = 0; i < N; i++) {
      if (i != 0) D[i][i - 1] = A;
      D[i][i] = B;
      if (i + 1 != N) D[i][i + 1] = C;
    }

    vec E(N);
    for (int i = 0; i < N; i++) {
      cin >> E[i];
    }

    D = mat_pow(D, T, M);
    E = mul(D, E, M);
    for (int i = 0; i < N; i++) {
      cout << E[i] << " \n"[i + 1 == N];
    }
  }

  return 0;
}

### 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 <bits/stdc++.h>

using namespace std;
using ll = long long;
using pii = pair<int, int>;

using vec = vector<int>;
using mat = vector<vec>;
 
vec mul(mat &A, vec &B, ll mod) {
  vec C(A.size());
  for (int i = 0; i < A.size(); i++) for (int j = 0; j < B.size(); j++) {
    (C[i] += A[i][j] * B[j]) %= mod;
  }
  return C;
}

mat mul(mat &A, mat &B, ll mod) {
  mat C(A.size(), vec(B[0].size()));
  for (int i = 0; i < A.size(); i++) for (int k = 0; k < B.size(); k++) for (int j = 0; j < B[0].size(); j++) {
    (C[i][j] += A[i][k] * B[k][j]) %= mod;
  }
  return C;
}
 
mat mat_pow(mat A, ll n, ll mod) {
  mat B(A.size(), vec(A.size()));
  for (int i = 0; i < A.size(); i++) {
    B[i][i] = 1;
  }
  while (n) {
    if (n & 1) B = mul(B, A, mod);
    A = mul(A, A, mod);
    n >>= 1;
  }
  return B;
}

int N, M, A, B, C, T;

int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(10);
  
  while (cin >> N >> M >> A >> B >> C >> T, N) {
    mat D(N, vec(N));
    for (int i = 0; i < N; i++) {
      if (i != 0) D[i][i - 1] = A;
      D[i][i] = B;
      if (i + 1 != N) D[i][i + 1] = C;
    }

    vec E(N);
    for (int i = 0; i < N; i++) {
      cin >> E[i];
    }

    D = mat_pow(D, T, M);
    E = mul(D, E, M);
    for (int i = 0; i < N; i++) {
      cout << E[i] << " \n"[i + 1 == N];
    }
  }

  return 0;
}

```