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 C (judge verdict: Wrong Answer)

#include <stdio.h>
#include <time.h>
int main(void){
	int a = 1, n;
	char c;
	
	while( scanf("%d%c %*c %*c %*c", &n, &c ), n != 0 ){
		if( c == ' ' ){
			clock_t c1 = clock();
			clock_t c2 = (clock_t)(a * 0.1 * CLOCKS_PER_SEC + c1);
			while( clock() < c2 );
			return 0;
		}
		++a;
	}
}