#include <bits/stdc++.h>
using namespace std;


int N,M,A,B,C,TT;

typedef vector<int> Arr;
typedef vector<Arr> Mat;


Mat mul(const Mat &a,const Mat &b){
	int n = a.size();
	Mat c(n,Arr(n,0));
	for(int i = 0 ; i < n ; i++)
		for(int j = 0 ; j < n ; j++)	
			for(int k = 0 ; k < n ; k++)
				c[i][j] = (c[i][j]+(a[i][k] * b[k][j])) % M;
	return c;
}
Mat E(int n){
	Mat c(n,Arr(n,0));
	for(int i = 0 ; i < n ; i++) c[i][i] = 1;
	return c;
}
Mat pow(Mat a,int k){
	Mat res = E(a.size());
	if( k == 0 ) return res;
	if(k&1) res = mul(res,a);
	res = mul(res,pow(mul(a,a),k/2));
	return res;
}

int main(){
	while(cin >> N >> M >> A >> B >> C >> TT && N){
		int S[50] = {};
		for(int i = 0 ; i < N ; i++) cin >> S[i];
		Mat mat(N,Arr(N,0));
		for(int i = 1 ; i < N ; i++) mat[i][i-1] = A;
		for(int i = 0 ; i < N ; i++) mat[i][i] = B;
		for(int i = 0 ; i+1 < N ; i++) mat[i][i+1] = C;

		Mat cal = pow(mat,TT);
		
		for(int i = 0 ; i < N ; i++){
			int ans = 0;
			for(int j = 0 ; j < N ; j++){
				ans += cal[i][j] * S[j];
			}
			ans %= M;
			cout << ans << (i+1==N?"\n":" ");
		}
	}
}
### Prompt

Create a solution in Cpp for the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;


int N,M,A,B,C,TT;

typedef vector<int> Arr;
typedef vector<Arr> Mat;


Mat mul(const Mat &a,const Mat &b){
	int n = a.size();
	Mat c(n,Arr(n,0));
	for(int i = 0 ; i < n ; i++)
		for(int j = 0 ; j < n ; j++)	
			for(int k = 0 ; k < n ; k++)
				c[i][j] = (c[i][j]+(a[i][k] * b[k][j])) % M;
	return c;
}
Mat E(int n){
	Mat c(n,Arr(n,0));
	for(int i = 0 ; i < n ; i++) c[i][i] = 1;
	return c;
}
Mat pow(Mat a,int k){
	Mat res = E(a.size());
	if( k == 0 ) return res;
	if(k&1) res = mul(res,a);
	res = mul(res,pow(mul(a,a),k/2));
	return res;
}

int main(){
	while(cin >> N >> M >> A >> B >> C >> TT && N){
		int S[50] = {};
		for(int i = 0 ; i < N ; i++) cin >> S[i];
		Mat mat(N,Arr(N,0));
		for(int i = 1 ; i < N ; i++) mat[i][i-1] = A;
		for(int i = 0 ; i < N ; i++) mat[i][i] = B;
		for(int i = 0 ; i+1 < N ; i++) mat[i][i+1] = C;

		Mat cal = pow(mat,TT);
		
		for(int i = 0 ; i < N ; i++){
			int ans = 0;
			for(int j = 0 ; j < N ; j++){
				ans += cal[i][j] * S[j];
			}
			ans %= M;
			cout << ans << (i+1==N?"\n":" ");
		}
	}
}
```