#include <iostream>
using namespace std;
int N, M, A, B, C, T, S[55], X[55][55], Y[55][55], Z[55][55];
void mul() {
	int R[55][55];
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < N; j++) {
			int sum = 0;
			for (int k = 0; k < N; k++) sum += X[i][k] * Y[k][j];
			R[i][j] = sum % M;
		}
	}
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < N; j++) X[i][j] = R[i][j];
	}
}
void power() {
	for (int i = 30; i >= 0; i--) {
		for (int j = 0; j < N; j++) {
			for (int k = 0; k < N; k++) Y[j][k] = X[j][k];
		}
		mul();
		if (T & (1 << i)) {
			for (int j = 0; j < N; j++) {
				for (int k = 0; k < N; k++) Y[j][k] = Z[j][k];
			}
			mul();
		}
	}
}
int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	while (cin >> N >> M >> A >> B >> C >> T, N) {
		for (int i = 0; i < N; i++) cin >> S[i];
		for (int i = 0; i < N; i++) {
			Z[i][i] = B;
			if (i != 0) Z[i][i - 1] = A;
			if (i + 1 != N) Z[i][i + 1] = C;
		}
		for (int i = 0; i < N; i++) {
			for (int j = 0; j < N; j++) X[i][j] = 0;
			X[i][i] = 1;
		}
		power();
		for (int i = 0; i < N; i++) {
			int ret = 0;
			for (int j = 0; j < N; j++) ret += X[i][j] * S[j];
			if (i) cout << ' ' << ret % M;
			else cout << ret % M;
		}
		cout << endl;
	}
}
### 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;
int N, M, A, B, C, T, S[55], X[55][55], Y[55][55], Z[55][55];
void mul() {
	int R[55][55];
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < N; j++) {
			int sum = 0;
			for (int k = 0; k < N; k++) sum += X[i][k] * Y[k][j];
			R[i][j] = sum % M;
		}
	}
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < N; j++) X[i][j] = R[i][j];
	}
}
void power() {
	for (int i = 30; i >= 0; i--) {
		for (int j = 0; j < N; j++) {
			for (int k = 0; k < N; k++) Y[j][k] = X[j][k];
		}
		mul();
		if (T & (1 << i)) {
			for (int j = 0; j < N; j++) {
				for (int k = 0; k < N; k++) Y[j][k] = Z[j][k];
			}
			mul();
		}
	}
}
int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	while (cin >> N >> M >> A >> B >> C >> T, N) {
		for (int i = 0; i < N; i++) cin >> S[i];
		for (int i = 0; i < N; i++) {
			Z[i][i] = B;
			if (i != 0) Z[i][i - 1] = A;
			if (i + 1 != N) Z[i][i + 1] = C;
		}
		for (int i = 0; i < N; i++) {
			for (int j = 0; j < N; j++) X[i][j] = 0;
			X[i][i] = 1;
		}
		power();
		for (int i = 0; i < N; i++) {
			int ret = 0;
			for (int j = 0; j < N; j++) ret += X[i][j] * S[j];
			if (i) cout << ' ' << ret % M;
			else cout << ret % M;
		}
		cout << endl;
	}
}
```