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

typedef vector<int> vec;
typedef vector<vec> mat;

mat mul(mat A, mat B, int M){
  if(A[0].size() != B.size())return mat();
  int n = A.size(), m = B[0].size();
  mat C(n,vec(m));

  for(int i=0;i<n;i++)
    for(int j=0;j<m;j++)
      for(int k=0;k<A[0].size();k++)
	C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % M;
  return C;
}

mat pow(mat A,int n,int M){
  mat res(A.size(),vec(A[0].size()));
  for(int i=0;i<A.size();i++)res[i][i] = 1;
  while(n>0){
    if(n&1)res = mul(res,A,M);
    A = mul(A,A,M);
    n >>= 1;
  }
  return res;
}

int main(){
  int n,m,a,b,c,t;

  while(scanf("%d%d%d%d%d%d",&n,&m,&a,&b,&c,&t),(n||m||a||b||c||t)){
    mat s(n,vec(1)),r;
    for(int i=0;i<n;i++)scanf("%d",&s[i][0]);
    
    mat A(n,vec(n));
    for(int i=0;i<n;i++){
      if(i)A[i][i-1] = a;
      A[i][i] = b;
      if(i!=n-1)A[i][i+1] = c;
    }

    r = mul(pow(A,t,m),s,m);
    for(int i=0;i<n;i++)printf("%d%c",r[i][0],((i==n-1)?'\n':' '));
  }
}