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 JAVA (judge verdict: Accepted)

import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.*;
import java.util.Calendar.Builder;



public class Main {
	static int INF = 2 << 27;
	static int N;
	static int M;
	static int A;
	static int B;
	static int C;
	static int T;
	static long[][] dp;
	public static void main(String[] args) {	
		//FastScanner sc = new FastScanner();
		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		while(true) {
			N = sc.nextInt();
			M = sc.nextInt();
			A = sc.nextInt();
			B = sc.nextInt();
			C = sc.nextInt();
			T = sc.nextInt();
			
			if(N == 0 && M == 0 && A == 0 && B == 0 && C == 0 && T == 0) break;
			long[] X = new long[N];
			for(int i = 0; i < N; i++) {
				X[i] = sc.nextLong();
			}
			
			long[][] matrix = new long[N][N];
			for(int i = 0; i < N; i++) {
				if(i - 1 >= 0) matrix[i][i - 1] = A;
				matrix[i][i] = B;
				if(i + 1 < N) matrix[i][i + 1] = C;
			}
			
			String TB = Integer.toBinaryString(T);
			long[][] mt = new long[N][N];
			for(int i = 0; i < N; i++) {
				mt[i][i] = 1;
			}
			for(int i = 0; i < TB.length(); i++) {
				if(TB.charAt(TB.length() - i - 1) == '1') {
					mt = mul(mt, matrix);
					mt = mod(mt);
				}
				matrix = mul(matrix,matrix);
				matrix = mod(matrix);
				
			}
			
			
			
			for(int i = 0; i < N; i++) {
				long ans = 0;
				for(int j = 0; j < N; j++) {
					ans = (mt[i][j] * X[j] + ans) % M;
				}
				if(i != 0) System.out.print(" " + ans);
				else System.out.print(ans);
			}
			System.out.println();
		}
	}
	static long[][] mul(long[][] a, long[][] b) {
		int n = a.length;
		long[][] ret = new long[n][n];
		for(int i = 0; i < n; i++) {
			for(int j = 0; j < n; j++) {
				for(int k = 0; k < n; k++) {
					ret[i][j] = (a[i][k] * b[k][j] + ret[i][j]);
				}
			}
		}
		return ret;
	}
	static long[][] mod(long[][] a) {
		long[][] ret = new long[a.length][a.length];
		for(int i = 0; i < a.length; i++) {
			for(int j = 0; j < a[i].length; j++) {
				ret[i][j] = a[i][j] % M;
			}
		}
		return ret;
	}
	
	static long[][] add(long[][] a, long[][] b) {
		int n = a.length;
		long[][] ret = new long[n][n];
		for(int i = 0; i < n; i++) {
			for(int j = 0; j < n; j++) {
				ret[i][j] = (a[i][j] + b[i][j]);
			}
		}
		return ret;
	}

}