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.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.Scanner;

public class Main {
	
	public static int[][] matrix_mul(final int n, final int mod, int[][] a, int[][] b){
		int[][] ret = new int[n][n];
		
		for(int i = 0; i < n; i++){
			for(int j = 0;j < n; j++){
				int sum = 0;
				
				for(int k = 0; k < n; k++){
					sum += (a[i][k]* b[k][j])  % mod;
					sum %= mod;
				}
				
				ret[i][j] = sum % mod;
			}
		}
		
		return ret;
	}
	
	public static int[][] matrix_pow(final int n, final int mod, final int t, int[][] ret){
		if(t == 1){
			return ret;
		}
		
		int[][] tmp = matrix_pow(n, mod, t / 2, ret);
		int[][] tmp_2 = matrix_mul(n, mod, tmp, tmp);
		
		if(t % 2 == 0){
			return tmp_2;
		}else{
			return matrix_mul(n, mod, tmp_2, ret);
		}
	}
	
	public static int[] vec_max_mul(final int n, final int mod, int[] array, int[][] matrix){
		int[] ret = new int[n];
		
		for(int i = 0; i < n; i++){
			int sum = 0;
			
			for(int j = 0;j < n; j++){
				sum += (array[j] * matrix[j][i]) % mod;
				sum %= mod;
			}
			
			ret[i] = sum;
		}
		
		return ret;
	}
	
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);

		while(true){
			final int n = sc.nextInt();
			final int m = sc.nextInt();
			final int a = sc.nextInt();
			final int b = sc.nextInt();
			final int c = sc.nextInt();
			final int t = sc.nextInt();
			
			if(n == 0 && m == 0 && a == 0 && b == 0 && c == 0 && t == 0){
				break;
			}
			
			int[] array = new int[n];
			for(int i = 0; i < n; i++){
				array[i] = sc.nextInt();
			}
			
			int[][] matrix = new int[n][n];
			for(int i = 0; i < n; i++){
				if(i != 0){
					matrix[i-1][i] = a;	
				}
				
				matrix[i][i] = b;
				
				if(i != n-1){
					matrix[i+1][i] = c;
				}
			}
			/*
			for(int[] tmp : matrix){
				System.out.println(Arrays.toString(tmp));
			}
			*/
			int[] ret = null;
			if(t == 0){
				ret = array;
			}else{
				ret = vec_max_mul(n, m, array, matrix_pow(n, m, t, matrix));
			}
			
			
			boolean first = true;
			for(int i = 0; i < n; i++){
				if(first){
					System.out.print(ret[i]);
					first = false;
				}else{
					System.out.print(" " + ret[i]);
				}
			}
			
			System.out.println();
		}
		
	}

}