#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;

vector<vector<int> > mul_mat(const vector<vector<int> > &a, const vector<vector<int> > &b, int m)
{
	int n = a.size();
	vector<vector<int> > ret;

	ret.resize(n);
	for(int i = 0; i < n; ++i)
		ret[i].resize(n);

	for(int i = 0; i < n; ++i) {
		for(int j = 0; j < n; ++j) {
			ret[i][j] = 0;
			for(int k = 0; k < n; ++k)
				ret[i][j] = (ret[i][j] + a[i][k] * b[k][j]) % m;
		}
	}

	return ret;
}

vector<vector<int> > pow_mat(const vector<vector<int> > &mat, int k, int m)
{
	int n = mat.size();
	vector<vector<int> > ret, x = mat;

	ret.resize(n);
	for(int i = 0; i < n; ++i)
		ret[i].resize(n, 0);

	for(int i = 0; i < n; ++i)
		ret[i][i] = 1;

	for(int bit = 0; bit < 31; ++bit, x = mul_mat(x, x, m)) {

		if((k & (1 << bit)) == 0)
			continue;
		ret = mul_mat(ret, x, m);
	}

	return ret;
}

int main()
{
	while(true) {

		int n, m, a, b, c, t;
		vector<int> s;

		scanf("%d%d%d%d%d%d", &n, &m, &a, &b, &c, &t);

		if(n == 0)
			break;

		s.resize(n);
		for(int i = 0; i < n; ++i)
			scanf("%d", &s[i]);

		vector<vector<int> > mat;

		mat.resize(n);
		for(int i = 0; i < n; ++i)
			mat[i].resize(n);

		for(int i = 0; i < n; ++i) {
			if(i - 1 >= 0) mat[i][i - 1] = a;
			if(i + 1 < n) mat[i][i + 1] = c;
			mat[i][i] = b;
		}

		vector<vector<int> > x = pow_mat(mat, t, m);
		vector<int> ans(n);

		for(int i = 0; i < n; ++i) {
			ans[i] = 0;
			for(int j = 0; j < n; ++j)
				ans[i] = (ans[i] + x[i][j] * s[j]) % m;
		}

		for(int i = 0; i < n; ++i)
			printf("%d%c", ans[i], i < n - 1 ? ' ' : '\n');
	}

	return 0;
}
### Prompt

Develop a solution in cpp to the problem described below:

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 <cstdio>
#include <vector>
#include <algorithm>
using namespace std;

vector<vector<int> > mul_mat(const vector<vector<int> > &a, const vector<vector<int> > &b, int m)
{
	int n = a.size();
	vector<vector<int> > ret;

	ret.resize(n);
	for(int i = 0; i < n; ++i)
		ret[i].resize(n);

	for(int i = 0; i < n; ++i) {
		for(int j = 0; j < n; ++j) {
			ret[i][j] = 0;
			for(int k = 0; k < n; ++k)
				ret[i][j] = (ret[i][j] + a[i][k] * b[k][j]) % m;
		}
	}

	return ret;
}

vector<vector<int> > pow_mat(const vector<vector<int> > &mat, int k, int m)
{
	int n = mat.size();
	vector<vector<int> > ret, x = mat;

	ret.resize(n);
	for(int i = 0; i < n; ++i)
		ret[i].resize(n, 0);

	for(int i = 0; i < n; ++i)
		ret[i][i] = 1;

	for(int bit = 0; bit < 31; ++bit, x = mul_mat(x, x, m)) {

		if((k & (1 << bit)) == 0)
			continue;
		ret = mul_mat(ret, x, m);
	}

	return ret;
}

int main()
{
	while(true) {

		int n, m, a, b, c, t;
		vector<int> s;

		scanf("%d%d%d%d%d%d", &n, &m, &a, &b, &c, &t);

		if(n == 0)
			break;

		s.resize(n);
		for(int i = 0; i < n; ++i)
			scanf("%d", &s[i]);

		vector<vector<int> > mat;

		mat.resize(n);
		for(int i = 0; i < n; ++i)
			mat[i].resize(n);

		for(int i = 0; i < n; ++i) {
			if(i - 1 >= 0) mat[i][i - 1] = a;
			if(i + 1 < n) mat[i][i + 1] = c;
			mat[i][i] = b;
		}

		vector<vector<int> > x = pow_mat(mat, t, m);
		vector<int> ans(n);

		for(int i = 0; i < n; ++i) {
			ans[i] = 0;
			for(int j = 0; j < n; ++j)
				ans[i] = (ans[i] + x[i][j] * s[j]) % m;
		}

		for(int i = 0; i < n; ++i)
			printf("%d%c", ans[i], i < n - 1 ? ' ' : '\n');
	}

	return 0;
}
```