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

Solution (in CPP):

Solution:
#include <iostream>
#include <vector>
using namespace std;
int main(int argc, char *argv[])
{
  for(;;) {
    int n, m, a, b, c, t;
    cin >> n >> m >> a >> b >> c >> t;
    if(n == 0) break;
    vector<int> ss;
    for(int i = 0; i < n; i++) {
      int s;
      cin >> s;
      ss.push_back(s);
    }
    // 0 < n <= 50
    // 0 < m <= 1000
    // 0 <= t <= 10^9 -> O(log t) is necessary
    // result = X^t A
    //    = [b c 0 ...; a b c 0 ...; ...; ... 0 a b]^T * [a0; a1; ...; an-1]

    vector<vector<int>> y(n, vector<int>(n, 0)); // to be X^(2^k)
    vector<vector<int>> z(n, vector<int>(n, 0)); // to be X^t
    for(int i = 0; i < n; i++) {
      for(int j = 0; j < n; j++) {
        if(i == j) {
          y[i][j] = b;
          z[i][j] = 1;
        } else if(i == j + 1) {
          y[i][j] = a;
        } else if(i == j - 1) {
          y[i][j] = c;
        }
      }
    }
    // z *= y
    auto mul = [](int n, int m, vector<vector<int>> &z, vector<vector<int>> &y) {
      int t[n][n];
      for(int i = 0; i < n; i++) {
        for(int j = 0; j < n; j++) {
          int c = 0;
          for(int k = 0; k < n; k++) {
            c = (c + (z[i][k] * y[k][j]) % m) % m;
          }
          t[i][j] = c;
        }
      }
      for(int i = 0; i < n; i++) {
        for(int j = 0; j < n; j++) {
          z[i][j] = t[i][j];
        }
      }
    };
    // computing X^t in O(log t) steps
    while(t != 0) {
      if(t & 1) { // z *= y
        mul(n, m, z, y);
      }
      mul(n, m, y, y);
      t >>= 1;
    }
    // compute and print the answer
    for(int i = 0; i < n; i++) {
      int r = 0;
      for(int k = 0; k < n; k++) {
        r = (r + (z[i][k] * ss[k]) % m) % m;
      }
      if(i > 0) cout << " ";
      cout << r;
    }
    cout << endl;
  }
  return 0;
}