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 C (judge verdict: Wrong Answer)

#include <stdio.h>

int gcd(int a,int b) {
	return b==0?a:gcd(b,a%b);
}

int main(void) {
	int p,n;
	int x,y,u,v;
	int i,nowb,nows,g;
	int left,right,mid;
	while(scanf("%d%d",&p,&n)==2 && (p|n)!=0) {
		x=n;y=1;u=1;v=n;
		for(i=1;i<=n;i++) {
			/* search x/y */
			left=1;right=n;
			while(left<=right) {
				mid=(left+right)/2;
				if((long long)i*i*p<mid*mid) {
					right=mid-1;
				} else {
					left=mid+1;
				}
			}
			nows=right+1;
			g=gcd(nows,i);
			nows/=g;nowb=i/g;
			if(0<nows && nows<=n && nows*y<x*nowb){x=nows;y=nowb;}
			/* search u/v */
			left=1;right=n;
			while(left<=right) {
				mid=(left+right)/2;
				if(mid*mid<(long long)i*i*p) {
					left=mid+1;
				} else {
					right=mid-1;
				}
			}
			nows=left-1;
			g=gcd(nows,i);
			nows/=g;nowb=i/g;
			if(0<nows && nows<=n && u*nowb<nows*v){u=nows;v=nowb;}
		}
		printf("%d/%d %d/%d\n",x,y,u,v);
	}
	return 0;
}