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>
using namespace std;
int main(void){
  for(;;){
    int N,M,A,B,C,T;
    cin >> N >> M >> A >> B >> C >> T;
    if(N == 0 && M == 0 && A == 0 && B == 0 && C == 0 && T == 0){ return 0; }
    int S[N+2];
    S[0] = 0; S[N+1] = 0;
    for(int i = 1 ; i <= N ; i++){cin >> S[i];}
    // matrix a
    int a[N+2][N+2];
    for(int i = 0 ; i <= N+1 ; i++){
      a[0][i] = 0; a[N+1][i] = 0;
    }
    for(int i = 1 ; i <= N ; i++){
      for(int j = 0 ; j <= N+1 ; j++){
	if(i == j+1)     {a[i][j] = A;}
	else if(i == j)  {a[i][j] = B;}
	else if(i == j-1){a[i][j] = C;}
	else             {a[i][j] = 0;}
      }
    }
    // culculate a^T mod M
    int  temp[N+2][N+2];
    int  ans [N+2][N+2];
    bool first = true;
    for(int i = 0 ; i <= N+1 ; i++){
      for(int j = 0 ; j <= N+1 ; j++){
	ans[i][j]=((i==j)?(1):(0));
      }
    }
    for(int p = 0 ; (1<<p) <= T ; p++){
      int tmp[N+2][N+2];
      if((T>>p) % 2 != 0){
	for(int i = 0 ; i <= N+1 ; i++){
	  for(int j = 0 ; j <= N+1 ; j++){
	    tmp[i][j] = 0;
	    for(int k = 0 ; k <= N+1 ; k++){
	      tmp[i][j] += ans[i][k] * a[k][j];
	    }
	  }
	}
	for(int i = 0 ; i <= N+1 ; i++){
	  for(int j = 0 ; j <= N+1 ; j++){
	    ans[i][j] = tmp[i][j] % M;
	  }
	}
      }
      for(int i = 0 ; i <= N+1 ; i++){
	for(int j = 0 ; j <= N+1 ; j++){
	  tmp[i][j] = 0;
	  for(int k = 0 ; k <= N+1 ; k++){
	    tmp[i][j] += a[i][k] * a[k][j];
	  }
	}
      }
      for(int i = 0 ; i <= N+1 ; i++){
	for(int j = 0 ; j <= N+1 ; j++){
	  a[i][j] = tmp[i][j] % M;
	}
      }
    }
    
    int ansv[N+2];
    for(int i = 0 ; i <= N+1 ; i++){
      ansv[i] = 0;
      for(int k = 0 ; k <= N+1 ; k++){
	ansv[i] += ans[i][k] * S[k];
      }
    }
    for(int i = 1 ; i <= N ; i++){
      cout << ansv[i] % M << ((i == N)?("\n"):(" "));
    }
  }
}