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<string>
#include<cstring>
#include<vector>
#include<set>
typedef long long ll;
using namespace std;
typedef vector<int> vi;
int n,m,a,b,c,t;
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;
}

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()
{
  while(cin >> n >> m >> a >> b >> c >> t && n)
    {
      Matrix B(n,vector<int>(1));
      for(int i=0;i<n;i++)
	{
	  cin >> B[i][0];
	}
      Matrix A(n,vector<int>(n));
      for(int i=0;i<n;i++)
	{
	  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);
      for(int i=0;i<n;i++)
	{
	  cout << A[i][0] << (i+1==n ? '\n':' ');
	}
    }
  return 0;
}