Problem:
Problem C: One-Dimensional Cellular Automaton

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.

Sample 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 for the Sample Input

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 C++ (judge verdict: Accepted)

#include <iostream>
#include <vector>
using namespace std;

vector< vector<int> > matrix_product(vector< vector<int> > &A, vector< vector<int> > &B, int M){
  int n = A.size(), m = B[0].size(), l = A[0].size();
  vector< vector<int> > ret(n,vector<int>(m));
  for(int i = 0; i < n; ++i){
    for(int j = 0; j < m; ++j){
      ret[i][j] = 0;
      for(int k = 0; k < l; ++k){
        ret[i][j] += (A[i][k]*B[k][j])%M;
        ret[i][j] %= M;
      }
    }
  }
  return ret;
}

vector< vector<int> > matrix_pow(vector< vector<int> > &P, int N, int M){
  vector< vector<int> > ret(P.size(),vector<int>(P.size(),0)), t = P;
  for(int i = 0; i < P.size(); ++i) ret[i][i] = 1;
  int k = 0;
  while((1<<k) <= N){
    if((N>>k)&1){
      ret = matrix_product(ret,t,M);
    }
    ++k;
    t = matrix_product(t,t,M);
  }
  return ret;
}

int main(){
  int N, M, A, B, C, T;
  while(cin >> N >> M >> A >> B >> C >> T, N){
    vector< vector<int> > S(N,vector<int>(1));
    for(int i = 0; i < N; ++i) cin >> S[i][0];
    vector< vector<int> > P(N,vector<int>(N,0));
    for(int i = 0; i < N; ++i) P[i][i] = B;
    for(int i = 1; i < N; ++i) P[i][i-1] = A;
    for(int i = 0; i < N-1; ++i) P[i][i+1] = C;
    vector< vector<int> > P_ = matrix_pow(P,T,M);
    vector< vector<int> > ans = matrix_product(P_,S,M);
    for(int i = 0; i < N; ++i){
      cout << ans[i][0];
      if(i < N-1) cout << " ";
    }
    cout << endl;
  }
  return 0;
}