#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
#define rep(i,n) for(int i=0;i<n;i++)

vector<vector<int>> multi(vector<vector<int>> m1, vector<vector<int>> m2,int M)
{
	vector<vector<int>> ans(m1.size(), vector<int>(m2[0].size(), 0));

	rep(i, m1.size()) {
		rep(j, m2[0].size()) {
			rep(k, m1[0].size()) {
				ans[i][j] += m1[i][k] * m2[k][j];
			}

			ans[i][j] %= M;
		}
	}

	return ans;
}

vector<vector<int>> mypow(vector<vector<int>> matrix, int n, int M)
{
	if (n == 0) {
		vector<vector<int>> I(matrix.size(), vector<int>(matrix.size(), 0));

		rep(i, matrix.size()) {
			I[i][i] = 1;
		}

		return I;
	}
	else if (n % 2 == 0) {
		return mypow(multi(matrix, matrix, M), n / 2, M);
	}
	else {
		return multi(mypow(multi(matrix, matrix, M), n / 2, M), matrix, M);
	}
}

int main()
{
	while (true) {
		int N, M, A, B, C, T;

		cin >> N >> M >> A >> B >> C >> T;

		if (N == 0 && M == 0 && A == 0 && B == 0 && C == 0 && T == 0) {
			break;
		}

		vector<int> t_zero(N);

		rep(i, N) {
			cin >> t_zero[i];
		}

		if (T == 0) {
			rep(i, N) {
				cout << t_zero[i];

				if (i == N - 1) {
					cout << endl;
				}
				else {
					cout << " ";
				}
			}

			continue;
		}

		vector<vector<int>> matrix(N, vector<int>(N, 0));

		rep(i, N) {
			if (i - 1 >= 0) {
				matrix[i][i - 1] = A;
			}

			matrix[i][i] = B;

			if (i + 1 < N) {
				matrix[i][i + 1] = C;
			}
		}

		matrix = mypow(matrix, T, M);

		vector<int> ans(N, 0);

		rep(i, N) {
			rep(j, N) {
				ans[i] += matrix[i][j] * t_zero[j];
			}

			ans[i] %= M;
		}

		rep(i, N) {
			cout << ans[i];

			if (i == N - 1) {
				cout << endl;
			}
			else {
				cout << " ";
			}
		}
	}


	return 0;
}
### Prompt

Generate 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>
#include <vector>
#include <algorithm>
using namespace std;
#define rep(i,n) for(int i=0;i<n;i++)

vector<vector<int>> multi(vector<vector<int>> m1, vector<vector<int>> m2,int M)
{
	vector<vector<int>> ans(m1.size(), vector<int>(m2[0].size(), 0));

	rep(i, m1.size()) {
		rep(j, m2[0].size()) {
			rep(k, m1[0].size()) {
				ans[i][j] += m1[i][k] * m2[k][j];
			}

			ans[i][j] %= M;
		}
	}

	return ans;
}

vector<vector<int>> mypow(vector<vector<int>> matrix, int n, int M)
{
	if (n == 0) {
		vector<vector<int>> I(matrix.size(), vector<int>(matrix.size(), 0));

		rep(i, matrix.size()) {
			I[i][i] = 1;
		}

		return I;
	}
	else if (n % 2 == 0) {
		return mypow(multi(matrix, matrix, M), n / 2, M);
	}
	else {
		return multi(mypow(multi(matrix, matrix, M), n / 2, M), matrix, M);
	}
}

int main()
{
	while (true) {
		int N, M, A, B, C, T;

		cin >> N >> M >> A >> B >> C >> T;

		if (N == 0 && M == 0 && A == 0 && B == 0 && C == 0 && T == 0) {
			break;
		}

		vector<int> t_zero(N);

		rep(i, N) {
			cin >> t_zero[i];
		}

		if (T == 0) {
			rep(i, N) {
				cout << t_zero[i];

				if (i == N - 1) {
					cout << endl;
				}
				else {
					cout << " ";
				}
			}

			continue;
		}

		vector<vector<int>> matrix(N, vector<int>(N, 0));

		rep(i, N) {
			if (i - 1 >= 0) {
				matrix[i][i - 1] = A;
			}

			matrix[i][i] = B;

			if (i + 1 < N) {
				matrix[i][i + 1] = C;
			}
		}

		matrix = mypow(matrix, T, M);

		vector<int> ans(N, 0);

		rep(i, N) {
			rep(j, N) {
				ans[i] += matrix[i][j] * t_zero[j];
			}

			ans[i] %= M;
		}

		rep(i, N) {
			cout << ans[i];

			if (i == N - 1) {
				cout << endl;
			}
			else {
				cout << " ";
			}
		}
	}


	return 0;
}
```