#include<iostream>
using namespace std;

typedef long long ll;

int main() {
	ll mat[50][50], _mat[50][50], s[50], _s[50];
	ll n, m, a, b, c, t;
	while(1) {
		cin >> n >> m >> a >> b >> c >> t;
		if(!n&&!m&&!a&&!b&&!c&&!t) break;
		s[0] = s[n+1] = 0;
		for(int i=0; i<n; ++i) cin >> s[i];
		for(int i=0; i<n; ++i) {
			for(int j=0; j<n; ++j) {
				if(i-1==j) mat[i][j] = a;
				else if(i==j) mat[i][j] = b;
				else if(i==j-1) mat[i][j] = c;
				else mat[i][j] = 0;
			}
		}
		while(t) {
			if(t%2) {
				for(int i=0; i<n; ++i) _s[i] = s[i];
				for(int i=0; i<n; ++i) {
					s[i] = 0;
					for(int j=0; j<n; ++j) s[i] = (s[i] + mat[i][j]*_s[j])%m;
				}
			}
			for(int i=0; i<n; ++i) for(int j=0; j<n; ++j) _mat[i][j] = mat[i][j];
			for(int i=0; i<n; ++i) for(int j=0; j<n; ++j) {
				mat[i][j] = 0;
				for(int k=0; k<n; ++k) {
					mat[i][j] = (mat[i][j] + _mat[i][k]*_mat[k][j])%m;
				}
			}
			t/=2;
		}
		for(int i=0; i<n; ++i) cout << (!i?"":" ") << s[i];
		cout << endl;
	}
	return 0;
}
### Prompt

Please formulate a CPP solution to 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<iostream>
using namespace std;

typedef long long ll;

int main() {
	ll mat[50][50], _mat[50][50], s[50], _s[50];
	ll n, m, a, b, c, t;
	while(1) {
		cin >> n >> m >> a >> b >> c >> t;
		if(!n&&!m&&!a&&!b&&!c&&!t) break;
		s[0] = s[n+1] = 0;
		for(int i=0; i<n; ++i) cin >> s[i];
		for(int i=0; i<n; ++i) {
			for(int j=0; j<n; ++j) {
				if(i-1==j) mat[i][j] = a;
				else if(i==j) mat[i][j] = b;
				else if(i==j-1) mat[i][j] = c;
				else mat[i][j] = 0;
			}
		}
		while(t) {
			if(t%2) {
				for(int i=0; i<n; ++i) _s[i] = s[i];
				for(int i=0; i<n; ++i) {
					s[i] = 0;
					for(int j=0; j<n; ++j) s[i] = (s[i] + mat[i][j]*_s[j])%m;
				}
			}
			for(int i=0; i<n; ++i) for(int j=0; j<n; ++j) _mat[i][j] = mat[i][j];
			for(int i=0; i<n; ++i) for(int j=0; j<n; ++j) {
				mat[i][j] = 0;
				for(int k=0; k<n; ++k) {
					mat[i][j] = (mat[i][j] + _mat[i][k]*_mat[k][j])%m;
				}
			}
			t/=2;
		}
		for(int i=0; i<n; ++i) cout << (!i?"":" ") << s[i];
		cout << endl;
	}
	return 0;
}
```