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 <bits/stdc++.h>
using namespace std;
typedef long long lli;
typedef lli elm;
typedef vector<elm> vec;
typedef vector<vec> mat;

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

mat getE(int n) {
  mat e(n, vec(n, 0));
  for(int i = 0; i < n; ++i) {
    e[i][i] = 1;
  }
  return e;
}

mat multiply(const mat &a, const mat &b) {
  if(a.size() == 0 || b.size() == 0 ||
     a[0].size() != b.size()) return mat();
  int s = a.size();
  int t = b.size();
  int u = b[0].size();
  mat c(s, vec(u, 0));
  for(int i = 0; i < s; ++i) {
    for(int j = 0; j < u; ++j) {
      for(int k = 0; k < t; ++k) {
        ( c[i][j] += a[i][k] * b[k][j] % M ) %= M;
      }
    }
  }
  return c;
}

mat pow(const mat &a, const lli &n) {
  if(n == 0) return getE(a.size());
  mat b = pow(multiply(a, a), n/2);
  if(n & 1) return multiply(b, a);
  else return b;
}

int main() {
  while(cin >> N >> M >> A >> B >> C >> T &&
        (N|M|A|B|C|T)) {
    mat S(N, vec(1));
    for(int i = 0; i < N; ++i) cin >> S[i][0];

    mat D(N, vec(N));
    for(int i = 0; i < N; ++i) {
      if(i-1 >= 0) D[i][i-1] = A;
      D[i][i] = B;
      if(i+1 < N) D[i][i+1] = C;
    }
    D = pow(D, T);
    S = multiply(D, S);
    for(int i = 0; i < N; ++i) {
      if(i) cout << " ";
      cout << S[i][0];
    }
    cout << endl;
  }
  return 0;
}