#include<bits/stdc++.h>
#define rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
typedef vector<int>vec;
typedef vector<vec>mat;

int n, m, a, b, c, t;
mat mul(mat&a, mat&b) {
	mat c(a.size(), vec(b[0].size()));
	rep(i, a.size())rep(j, b[0].size())rep(k, b.size())
		(c[i][j] += a[i][k] * b[k][j]) %= m;
	return c;
}
mat ppow(mat&a, int b) {
	mat res(a.size(), vec(a.size()));
	rep(i, a.size())res[i][i] = 1;
	while (b) {
		if (b & 1)res = mul(res, a);
		a = mul(a, a);
		b >>= 1;
	}
	return res;
}
int main() {
	while (scanf("%d%d%d%d%d%d", &n, &m, &a, &b, &c, &t), n) {
		mat A(n, vec(n)), B(n, vec(1));
		rep(i, n)scanf("%d", &B[i][0]);
		rep(i, n)rep(j, n) {
			if (j == i - 1)A[i][j] = a;
			if (j == i)A[i][j] = b;
			if (j == i + 1)A[i][j] = c;
		}
		A = ppow(A, t);
		A = mul(A, B);
		rep(i, n) {
			if (i)printf(" ");
			printf("%d", A[i][0]);
		}
		puts("");
	}
}
### Prompt

Please provide a CPP coded solution 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<bits/stdc++.h>
#define rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
typedef vector<int>vec;
typedef vector<vec>mat;

int n, m, a, b, c, t;
mat mul(mat&a, mat&b) {
	mat c(a.size(), vec(b[0].size()));
	rep(i, a.size())rep(j, b[0].size())rep(k, b.size())
		(c[i][j] += a[i][k] * b[k][j]) %= m;
	return c;
}
mat ppow(mat&a, int b) {
	mat res(a.size(), vec(a.size()));
	rep(i, a.size())res[i][i] = 1;
	while (b) {
		if (b & 1)res = mul(res, a);
		a = mul(a, a);
		b >>= 1;
	}
	return res;
}
int main() {
	while (scanf("%d%d%d%d%d%d", &n, &m, &a, &b, &c, &t), n) {
		mat A(n, vec(n)), B(n, vec(1));
		rep(i, n)scanf("%d", &B[i][0]);
		rep(i, n)rep(j, n) {
			if (j == i - 1)A[i][j] = a;
			if (j == i)A[i][j] = b;
			if (j == i + 1)A[i][j] = c;
		}
		A = ppow(A, t);
		A = mul(A, B);
		rep(i, n) {
			if (i)printf(" ");
			printf("%d", A[i][0]);
		}
		puts("");
	}
}
```