Problem:
Problem A: Rational Irrationals

Rational numbers are numbers represented by ratios of two integers. For a prime number p, one of the elementary theorems in the number theory is that there is no rational number equal to √p. Such numbers are called irrational numbers. It is also known that there are rational numbers arbitrarily close to √p

Now, given a positive integer n, we define a set Qn of all rational numbers whose elements are represented by ratios of two positive integers both of which are less than or equal to n. For example, Q4 is a set of 11 rational numbers {1/1, 1/2, 1/3, 1/4, 2/1, 2/3, 3/1, 3/2, 3/4, 4/1, 4/3}. 2/2, 2/4, 3/3, 4/2 and 4/4 are not included here because they are equal to 1/1, 1/2, 1/1, 2/1 and 1/1, respectively.

Your job is to write a program that reads two integers p and n and reports two rational numbers x / y and u / v, where u / v < √p < x / y and there are no other elements of Qn between u/v and x/y. When n is greater than √p, such a pair of rational numbers always exists.

Input

The input consists of lines each of which contains two positive integers, a prime number p and an integer n in the following format.

p n

They are separated by a space character. You can assume that p and n are less than 10000, and that n is greater than √p. The end of the input is indicated by a line consisting of two zeros.

Output

For each input line, your program should output a line consisting of the two rational numbers x / y and u / v (x / y > u / v) separated by a space character in the following format.

x/y u/v

They should be irreducible. For example, 6/14 and 15/3 are not accepted. They should be reduced to 3/7 and 5/1, respectively.

Sample Input

2 5
3 10
5 100
0 0

Output for the Sample Input

3/2 4/3
7/4 5/3
85/38 38/17

Solution in JAVA (judge verdict: Accepted)

import java.util.*;
import java.awt.geom.*;

public class Main{	
	private void doit(){
		Scanner sc = new Scanner(System.in);
		while(true){
			int p = sc.nextInt();
			int n = sc.nextInt();
			if((p|n) == 0) break;
			double sqrtP = Math.sqrt(p);
			
			double nearMax = Integer.MAX_VALUE;
			int maxA = -1, maxB = -1;
			double nearMin = 0.0;
			int minA = -1, minB = -1;
			for(int i = 1; i <= n; i++){
				
				double temp = i * sqrtP;
				double temp2 = Math.ceil(temp);
				double temp1 = Math.floor(temp);
				double temp1i = temp1 / i;  // small
				double temp2i = temp2 / i;  //big
				
				if((int)temp1 <= n  && temp1i > nearMin){
					nearMin = temp1i;
					minA = (int) temp1;
					minB = i;
				}
				
				if((int) temp2 <= n && temp2i < nearMax){
					nearMax = temp2i;
					maxA = (int) temp2;
					maxB = i;
				}
			}
			System.out.println(maxA + "/" + maxB + " " + minA +"/"+minB);
		}
	}
	
	public static void main(String[] args) {
		new Main().doit();
	}
}