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 <bits/stdc++.h>
 
using namespace std;
 
typedef vector<int> vec;
typedef vector<vec> mat;
typedef long long ll;
 
int N,M,T;
 
mat mul(mat &A,mat &B){
    mat C(A.size(),vec(B[0].size()));
    for(int i = 0 ; i < (int)A.size() ; i++){
	for(int k = 0 ; k < (int)B.size() ; k++){
	    for(int j = 0 ; j < (int)B[0].size() ; j++){
		C[i][j] = (C[i][j] + A[i][k]*B[k][j]) % M;
	    }
	}
    }
    return C;
}
 
mat pow(mat A,ll n){
    mat B(A.size(),vec(A.size()));
    for(int i = 0 ; i < (int)A.size() ; i++){
	B[i][i] = 1;
    }
    while(n > 0){
	if(n & 1) B = mul(B,A);
	A = mul(A,A);
	n >>= 1;
    }
    return B;
}
 
int main(){
    int a[3];
    while(cin >> N >> M >> a[0] >> a[1] >> a[2] >> T, N){
	mat S(N,vec(1));
	for(int i = 0 ; i < N ; i++){
	    cin >> S[i][0];
	}
	mat A(N,vec(N));
	for(int i = 0 ; i < N ; i++){
	    for(int j = 0 ; j < N ; j++){
		A[i][j] = 0;
	    }
	    for(int j = i-1, k = 0 ; j < i+2 ; j++, k++){
		if(j < 0 || j >= N) continue;
		A[i][j] = a[k];
	    }
	}
	mat B = pow(A,T);
	mat C = mul(B,S);
	for(int i = 0 ; i < N ; i++){
	    if(i > 0) cout << " ";
	    cout << C[i][0];
	}
	cout << endl;
    }
    return 0;
}