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>

#define REP(i, x, n) for(int i = x; i < (int)(n); i++)
#define rep(i, n) REP(i, 0, n)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define F first
#define S second
#define mp make_pair
#define pb push_back

using namespace std;

typedef long long int lli;
typedef pair<int, int> Pii;

typedef vector<vector<int> > Matrix;

Matrix mul(const Matrix& A, const Matrix & B, const int mod) {
  Matrix C(A.size(), vector<int>(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] = (C[i][j] + A[i][k] * B[k][j]) % mod;
      }
    }
  }

  return C;
}

// A^n
Matrix pow(Matrix A, int n, int mod) {
  Matrix B(A.size(), vector<int>(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 main() {
  int n, m, a, b, c, t;
  
  while(cin >> n >> m >> a >> b >> c >> t) {
    if((n | m | a | b | c | t) == 0) break;

    Matrix B(n, vector<int>(1));
    rep(i, n) cin >> B[i][0];

    Matrix A(n, vector<int>(n));
    rep(i, n) {
      if(i - 1 >= 0) A[i][i - 1] = a;
      A[i][i] = b;
      if(i + 1 < n) A[i][i + 1] = c;
    }

    A = pow(A, t, m);
    A = mul(A, B, m);
    
    rep(i, n) {
      cout << A[i][0] << (i + 1 == n ? '\n' : ' ');
    }
  }

  return 0;
}