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

typedef vector<vector<int>> Matrix;

int n, m, a, b, c, t;
Matrix S, A;

Matrix mul(const Matrix &A, const Matrix &B)
{
	Matrix res(A.size(), vector<int>(B[0].size(), 0));
	
	for (int i = 0; i < A.size(); i++){
		for (int j = 0; j < B[0].size(); j++){
			for (int k = 0; k < A[0].size(); k++){
				res[i][j] += A[i][k] * B[k][j];
				res[i][j] %= m;
			}
		}
	}
	
	return res;
}

Matrix pow(const Matrix &A, int p)
{
	Matrix res(A.size(), vector<int>(A.size(), 0));
	for (int i = 0; i < A.size(); i++){
		res[i][i] = 1;
	}
	
	Matrix B = A;
	while (p){
		if (p & 1) res = mul(res, B);
		B = mul(B, B);
		p >>= 1;
	}
	
	return res;
}

int main()
{
	while (cin >> n >> m >> a >> b >> c >> t, n){
		A.assign(n, vector<int>(n, 0));
		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;
		}
		
		S.assign(n, vector<int>(1));
		for (int i = 0; i < n; i++){
			cin >> S[i][0];
		}
		
		Matrix res = mul(pow(A, t), S);
		for (int i = 0; i < n; i++){
			cout << res[i][0] << "\n "[i < n - 1];
		}
	}
}
### Prompt

Your challenge is to write 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 <bits/stdc++.h>
using namespace std;

typedef vector<vector<int>> Matrix;

int n, m, a, b, c, t;
Matrix S, A;

Matrix mul(const Matrix &A, const Matrix &B)
{
	Matrix res(A.size(), vector<int>(B[0].size(), 0));
	
	for (int i = 0; i < A.size(); i++){
		for (int j = 0; j < B[0].size(); j++){
			for (int k = 0; k < A[0].size(); k++){
				res[i][j] += A[i][k] * B[k][j];
				res[i][j] %= m;
			}
		}
	}
	
	return res;
}

Matrix pow(const Matrix &A, int p)
{
	Matrix res(A.size(), vector<int>(A.size(), 0));
	for (int i = 0; i < A.size(); i++){
		res[i][i] = 1;
	}
	
	Matrix B = A;
	while (p){
		if (p & 1) res = mul(res, B);
		B = mul(B, B);
		p >>= 1;
	}
	
	return res;
}

int main()
{
	while (cin >> n >> m >> a >> b >> c >> t, n){
		A.assign(n, vector<int>(n, 0));
		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;
		}
		
		S.assign(n, vector<int>(1));
		for (int i = 0; i < n; i++){
			cin >> S[i][0];
		}
		
		Matrix res = mul(pow(A, t), S);
		for (int i = 0; i < n; i++){
			cout << res[i][0] << "\n "[i < n - 1];
		}
	}
}
```