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++14 (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

vector<vector<int> > mat_dot(vector<vector<int> > first, vector<vector<int> > second, int mod) {
	int n = first.size();
	vector<vector<int> > res(n, vector<int>(n, 0));
	for (int i = 0; i < n; ++i) {
		for (int j = 0; j < n; ++j) {
			for (int k = 0; k < n; ++k) {
				res[i][j] += first[i][k] * second[k][j];
				res[i][j] %= mod;
			}
		}
	}
	return res;
}

vector<vector<int> > mat_pow(vector<vector<int> > mat, int exp, int mod) {
	int n = mat.size();
	vector<vector<int> > res(n, vector<int>(n, 0));
	for (int i = 0; i < n; ++i) res[i][i] = 1;
	int cnt = 0;
	while ((1<<cnt) <= exp) {
		if ((exp>>cnt) & 1) res = mat_dot(res, mat, mod);
		mat = mat_dot(mat, mat, mod);
		cnt++;
	}
	return res;
}

int main() {
	while (1) {
		int n, m, a, b, c, t;
		cin >> n >> m >> a >> b >> c >> t;
		if (n == 0) break;
		vector<int> s(n);
		for (int i = 0; i < n; ++i) cin >> s[i];
		vector<vector<int> > mt(n, vector<int>(n));
		for (int i = 0; i < n; ++i) {
			if (i > 0) mt[i][i-1] = a;
			mt[i][i] = b;
			if (i < n-1) mt[i][i+1] = c;
		}
		mt = mat_pow(mt, t, m);
		vector<int> ans(n);
		for (int i = 0; i < n; ++i) {
			for (int j = 0; j < n; ++j) {
				ans[i] += s[j] * mt[i][j];
				ans[i] %= m;
			}
			cout << (i ? " " : "") << ans[i];
		}
		cout << "\n";
	}
	return 0;
}