Problem:
Problem B: Cleaning Robot

Dr. Asimov, a robotics researcher, loves to research, but hates houseworks and his house were really dirty. So, he has developed a cleaning robot.

As shown in the following figure, his house has 9 rooms, where each room is identified by an alphabet:

The robot he developed operates as follows:

If the battery runs down, the robot stops.

If not so, the robot chooses a direction from four cardinal points with the equal probability, and moves to the room in that direction. Then, the robot clean the room and consumes 1 point of the battery.

However, if there is no room in that direction, the robot does not move and remains the same room. In addition, there is a junk room in the house where the robot can not enter, and the robot also remains when it tries to enter the junk room. The robot also consumes 1 point of the battery when it remains the same place.

A battery charger for the robot is in a room. It would be convenient for Dr. Asimov if the robot stops at the battery room when its battery run down.

Your task is to write a program which computes the probability of the robot stopping at the battery room.

Input

The input consists of several datasets. Each dataset consists of:

n
s t b

n is an integer that indicates the initial battery point. s, t, b are alphabets respectively represent the room where the robot is initially, the battery room, and the junk room.

The input ends with a dataset which consists of single 0 for n. Your program should not output for this dataset.

Output

For each dataset, print the probability as a floating point number in a line. The answer may not have an error greater than 0.000001.

Constraints

Judge data includes at most 100 data sets.

n ≤ 15

s, t, b are distinct.

Sample Input

1
E A C
1
E B C
2
E A B
0

Output for the Sample Input

0.00000000
0.25000000
0.06250000

Solution in JAVA (judge verdict: Accepted)

import java.io.*;

public class Main {
	public static void main(String[] args) {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		try {
			//initialize
			double dp[][][][] = new double[16][9][9][9];
			for(int i=0;i<16;i++){
				for(int j=0;j<9;j++){
					for(int k=0;k<9;k++){
						for(int l=0;l<9;l++){
							dp[i][j][k][l] = -1.0;
						}
					}
				}
			}
			String line;
			while(true){
				/* input */
				line = br.readLine();

				int n = Integer.parseInt(line);

				if(n==0) break;

				line = br.readLine();

				char s = line.split(" ")[0].charAt(0), t = line.split(" ")[1].charAt(0), b = line.split(" ")[2].charAt(0);

				/* processing */
				

				double ans = solve(dp, n,s-'A',t-'A',b-'A');
				/* output */
				System.out.println(ans);
			}
		} catch (IOException e) {
			e.printStackTrace();
		}
	}

	public static double solve(double[][][][] dp, int n, int s, int t, int b){
		if(dp[n][s][t][b]>=0) return dp[n][s][t][b];
		if(n==0){
			if(s==t){
				dp[n][s][t][b] = 1.0;
				return 1.0;
			} else {
				dp[n][s][t][b] = 0.0;
				return 0.0;
			}
		} else {
			double ret = 0;
			if(dp[n-1][s>2&&b!=s-3?s-3:s][t][b]>=0){
				ret += 0.25*dp[n-1][s>2&&b!=s-3?s-3:s][t][b];
			} else {
				dp[n-1][s>2&&b!=s-3?s-3:s][t][b] = solve(dp,n-1,s>2&&b!=s-3?s-3:s,t,b); 
				ret += 0.25*dp[n-1][s>2&&b!=s-3?s-3:s][t][b];
			}
			if(dp[n-1][s%3!=2&&b!=s+1?s+1:s][t][b]>=0){
				ret += 0.25*dp[n-1][s%3!=2&&b!=s+1?s+1:s][t][b];
			} else {
				dp[n-1][s%3!=2&&b!=s+1?s+1:s][t][b] = solve(dp,n-1,s%3!=2&&b!=s+1?s+1:s,t,b); 
				ret += 0.25*dp[n-1][s%3!=2&&b!=s+1?s+1:s][t][b];
			}
			if(dp[n-1][s<6&&b!=s+3?s+3:s][t][b]>=0){
				ret += 0.25*dp[n-1][s<6&&b!=s+3?s+3:s][t][b];
			} else {
				dp[n-1][s<6&&b!=s+3?s+3:s][t][b] = solve(dp,n-1,s<6&&b!=s+3?s+3:s,t,b);
				ret += 0.25*dp[n-1][s<6&&b!=s+3?s+3:s][t][b];
			}
			if(dp[n-1][s%3!=0&&b!=s-1?s-1:s][t][b]>=0){
				ret += 0.25*dp[n-1][s%3!=0&&b!=s-1?s-1:s][t][b];
			} else {
				dp[n-1][s%3!=0&&b!=s-1?s-1:s][t][b] =solve(dp,n-1,s%3!=0&&b!=s-1?s-1:s,t,b);
				ret += 0.25*dp[n-1][s%3!=0&&b!=s-1?s-1:s][t][b];
			}
			
			dp[n][s][t][b] = ret;
			return ret;
		}
	}
}