Problem:
Problem C: One-Dimensional Cellular Automaton

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.

Sample 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 for the Sample Input

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

Solution in C (judge verdict: Accepted)

// AOJ 1327 One-Dimensional Cellular Automaton
// 2017.10.25 bal4u
// 2018.2.9

#include <stdio.h>
#include <string.h>

int s[2][52], d[32][52][52];

#define getchar_unlocked()  getchar()
int in()
{
	int n = 0;
	int c = getchar_unlocked();
	do n = (n<<3)+(n<<1) + (c & 0xf), c = getchar_unlocked();
	while (c >= '0');
	return n;
}

int main()
{
	int n, m, a, b, c, t;
	int i, j, k, l, k1, k2;

	while (n = in()) {
		m = in(), a = in(), b = in(), c = in(), t = in();
		memset(d, 0, sizeof(d));
		for (i = 1; i <= n; i++) s[0][i] = in();
		for (i = 1; i <= n; i++) d[0][i][i+1] = a, d[0][i][i] = b, d[0][i][i-1] = c;
		for (k = 1; (1<<k) <= t; k++) for (i = 1; i <= n; i++) for (j = 1; j <= n; j++) {
			for (l = 1; l <= n; l++) d[k][i][j] += d[k-1][i][l] * d[k-1][l][j];
			d[k][i][j] %= m;
		}

		for (k1 = 0, k2 = 1, k = 0; (1<<k) <= t; k++) if (t & (1<<k)) {
			memset(s[k2], 0, sizeof(s[k2]));
			for (j = 1; j <= n; j++) {
				for (i = 1; i <= n; i++) s[k2][i] += s[k1][j] * d[k][j][i];
			}
			for (i = 1; i <= n; i++) s[k2][i] %= m;
			k1 = k2, k2 = !k2;
		}
		printf("%d", s[k1][1]);	for (i = 2; i <= n; i++) printf(" %d", s[k1][i]);
		putchar('\n');
	}
	return 0;
}