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 <vector>
#include <algorithm>
#include <cmath>
using namespace std;
const int inf = 1e9;

typedef long long int lli;
typedef vector<vector<lli> > Matrix;
bool mul(Matrix &a, Matrix &b, Matrix &c){ //A*B=C
	if(a[0].size() != b.size()) return false;
	int n = a.size();
	int m = b.size();
	int l = b[0].size();
	Matrix ret(n, vector<lli>(l,0));
	for(int i=0; i<n; i++){
		for(int j=0; j<l; j++){
			for(int k=0; k<m; k++){
				ret[i][j] += a[i][k]*b[k][j];
			}
		}
	}
	c = ret;
	return true;
}

int main(){
	while(1){
		int n,m,a,b,c,t;
		cin >> n >> m >> a >> b >> c >> t;
		if(n==0) break;
		
		Matrix ini(n, vector<lli>(1));
		for(int i=0; i<n; i++){
			cin >> ini[i][0];
		}
		
		Matrix r(n, vector<lli>(n));
		Matrix mat(n, vector<lli>(n, 0));
		for(int i=0; i<n; i++){
			r[i][i] = b;
			mat[i][i]=1;
			if(i!=0) r[i][i-1] = a;
			if(i!=n-1) r[i][i+1] = c;
		}
		for(int i=log2(t); i>=0; i--){
			mul(mat, mat, mat);
			if((t&1<<i) != 0){
				mul(mat, r, mat);
			}
			for(int j=0; j<n; j++){
				for(int k=0; k<n; k++){
					mat[j][k] = mat[j][k]%m;
				}
			}
		}
		
		mul(mat, ini, ini);
		for(int i=0; i<n-1; i++){
			cout << ini[i][0]%m << " ";
		}
		cout << ini[n-1][0]%m << endl;
	}
	return 0;
}