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 <algorithm>
#include <vector>

#define phb push_back

using namespace std;

typedef vector< int > Row;
typedef vector< Row > Matrix;

int N, M, A, B, C, T;
int S[50], res[50];

void read();
void soups_on();
Matrix mul(Matrix &, Matrix &);

int main() {
  while (cin >> N >> M >> A >> B >> C >> T, N)
    read(), soups_on();

  return 0;
}

void read() {
  for (int i = 0; i < N; ++i)
    cin >> S[i];
}

void soups_on() {
  Matrix base, mtrx;

  mtrx = base = Matrix(N, Row(N, 0));
  for (int i = 0; i < N; ++i)
    mtrx[i][i] = 1;

  for (int i = 0; i < N; ++i) {
    if (i > 0) base[i][i - 1] = A;
    base[i][i] = B;
    if (i + 1 < N) base[i][i + 1] = C;
  }

  for (int q = 30; q >= 0; --q) {
    mtrx = mul(mtrx, mtrx);
    if (T & 1 << q)
      mtrx = mul(mtrx, base);
  }

  fill(&res[0], &res[N], 0);
  for (int i = 0; i < N; ++i)
    for (int j = 0; j < N; ++j)
      res[i] += mtrx[i][j] * S[j], res[i] %= M;

  for (int i = 0; i < N; ++i) {
    if (i > 0) cout << " ";
    cout << res[i];
  }

  cout << "\n";
}

Matrix mul(Matrix &ma, Matrix &mb) {
  Matrix mr(N, Row(N, 0));
  for (int i = 0; i < N; ++i)
    for (int j = 0; j < N; ++j)
      for (int k = 0; k < N; ++k)
        mr[i][j] += ma[i][k] * mb[k][j], mr[i][j] %= M;
  return mr;
}