#include<iostream>
#include<vector>
using namespace std;
struct Matrix { vector<vector<long long>>x; };
void resize_(Matrix &a, int H, int W) {
	a.x.resize(H);
	for (int i = 0; i < H; i++)a.x[i].resize(W);
	for (int i = 0; i < H; i++) { for (int j = 0; j < W; j++)a.x[i][j] = 0; }
}
Matrix multiple(Matrix a, Matrix b, long long m) {
	Matrix c; resize_(c, a.x.size(), b.x[0].size());
	for (int i = 0; i < c.x.size(); i++) {
		for (int j = 0; j < c.x[0].size(); j++) {
			for (int k = 0; k < b.x.size(); k++) {
				c.x[i][j] += a.x[i][k] * b.x[k][j];
				c.x[i][j] %= m;
			}
		}
	}
	return c;
}
Matrix power(Matrix a, long long r, long long m) {
	Matrix G[64], H; G[0] = a; int cnt = 0; H = a;
	for (int i = 0; i < 63; i++) {
		G[i + 1] = multiple(G[i], G[i], m);
	}
	for (int i = 0; i < a.x.size(); i++) { for (int j = 0; j < a.x[0].size(); j++) { H.x[i][j] = 0; } }
	for (int i = 0; i < a.x.size(); i++) { H.x[i][i] = 1; }
	for (long long i = 0; i < 63; i++) {
		if ((r / (1LL << i)) % 2 == 0)continue;
		H = multiple(H, G[i], m);
	}
	return H;
}
int main() {
	while (true) {
		long long 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;
		a %= m; b %= m; c %= m;
		Matrix M1; resize_(M1, n, 1);
		for (int i = 0; i < n; i++) { cin >> M1.x[i][0]; }
		Matrix M2; resize_(M2, n, n);
		for (int i = 0; i < n; i++) {
			if (i > 0)M2.x[i][i - 1] = a;
			M2.x[i][i] = b;
			if (i < n - 1)M2.x[i][i + 1] = c;
		}
		Matrix M3 = power(M2, t, m);
		Matrix M4 = multiple(M3, M1, m);
		for (int i = 0; i < n; i++) { if (i)cout << ' '; cout << M4.x[i][0]; }cout << endl;
	}
	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<iostream>
#include<vector>
using namespace std;
struct Matrix { vector<vector<long long>>x; };
void resize_(Matrix &a, int H, int W) {
	a.x.resize(H);
	for (int i = 0; i < H; i++)a.x[i].resize(W);
	for (int i = 0; i < H; i++) { for (int j = 0; j < W; j++)a.x[i][j] = 0; }
}
Matrix multiple(Matrix a, Matrix b, long long m) {
	Matrix c; resize_(c, a.x.size(), b.x[0].size());
	for (int i = 0; i < c.x.size(); i++) {
		for (int j = 0; j < c.x[0].size(); j++) {
			for (int k = 0; k < b.x.size(); k++) {
				c.x[i][j] += a.x[i][k] * b.x[k][j];
				c.x[i][j] %= m;
			}
		}
	}
	return c;
}
Matrix power(Matrix a, long long r, long long m) {
	Matrix G[64], H; G[0] = a; int cnt = 0; H = a;
	for (int i = 0; i < 63; i++) {
		G[i + 1] = multiple(G[i], G[i], m);
	}
	for (int i = 0; i < a.x.size(); i++) { for (int j = 0; j < a.x[0].size(); j++) { H.x[i][j] = 0; } }
	for (int i = 0; i < a.x.size(); i++) { H.x[i][i] = 1; }
	for (long long i = 0; i < 63; i++) {
		if ((r / (1LL << i)) % 2 == 0)continue;
		H = multiple(H, G[i], m);
	}
	return H;
}
int main() {
	while (true) {
		long long 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;
		a %= m; b %= m; c %= m;
		Matrix M1; resize_(M1, n, 1);
		for (int i = 0; i < n; i++) { cin >> M1.x[i][0]; }
		Matrix M2; resize_(M2, n, n);
		for (int i = 0; i < n; i++) {
			if (i > 0)M2.x[i][i - 1] = a;
			M2.x[i][i] = b;
			if (i < n - 1)M2.x[i][i + 1] = c;
		}
		Matrix M3 = power(M2, t, m);
		Matrix M4 = multiple(M3, M1, m);
		for (int i = 0; i < n; i++) { if (i)cout << ' '; cout << M4.x[i][0]; }cout << endl;
	}
	return 0;
}
```