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: Time Limit Exceeded)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
	final static byte nexts[][] = {
			// 東,西,南,北
			{ 1, -1, 3, -1 },// 0:A
			{ 2, 0, 4, -1 }, // 1:B
			{ -1, 1, 5, -1 },// 2:C
			{ 4, -1, 6, 0 }, // 3:D
			{ 5, 3, 7, 1 }, // 4:E
			{ -1, 4, 8, 2 }, // 5:F
			{ 7, -1, -1, 3 },// 6:G
			{ 8, 6, -1, 4 }, // 7:H
			{ -1, 7, -1, 5 } // 8:I
	};
	final static byte distances[][] = {
			// A,B, C, D, E, F, G, H, I
			{ 0, 1, 2, 1, 2, 3, 2, 3, 4 },// 0:A
			{ 1, 0, 1, 2, 1, 2, 3, 2, 3 },// 1:B
			{ 2, 1, 0, 3, 2, 1, 4, 3, 2 },// 2:C
			{ 1, 2, 3, 0, 1, 2, 1, 2, 3 },// 3:D
			{ 2, 1, 2, 1, 0, 1, 2, 1, 2 },// 4:E
			{ 3, 2, 1, 2, 1, 0, 3, 2, 1 },// 5:F
			{ 2, 3, 4, 1, 2, 3, 0, 1, 2 },// 6:G
			{ 3, 2, 3, 2, 1, 2, 1, 0, 1 },// 7:H
			{ 4, 3, 2, 3, 2, 1, 2, 1, 0 } // 8:I
	};

	static byte charger;
	static byte storeroom;
	static int probability;

	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String line = "";

		while ((line = br.readLine()) != null && !line.isEmpty()) {
			byte battery;
			byte start;

			battery = Byte.parseByte(line);
			if (battery == 0) {
				break;
			}

			line = br.readLine();
			start = (byte) (line.toCharArray()[0] - 'A');
			charger = (byte) (line.toCharArray()[2] - 'A');
			storeroom = (byte) (line.toCharArray()[4] - 'A');

			probability = 0;
			clean(start, battery, 1 << battery * 2);
			System.out.println(probability * Math.pow(4, -battery));
		}
	}

	static void clean(byte current, int battery, int p) {
		if (battery == 0) {
			if (current == charger) {
				probability += p;
			}
		} else if (distance(current, charger) <= battery) {
			battery--;
			p >>= 2;
			for (byte n : next(current)) {
				if (n < 0 || n == storeroom) {
					clean(current, battery, p);
				} else {
					clean(n, battery, p);
				}
			}
		}
	}

	static byte[] next(byte current) {
		return nexts[current];
	}

	static byte distance(byte current, byte charger) {
		return distances[current][charger];
	}
}