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 Ruby (judge verdict: Accepted)

while true
    p, n = gets.chomp.split.map(&:to_i)
    break if [n, p] == [0, 0]
    rootp = Math.sqrt(p)
    left, right = Rational(rootp.floor, 1), Rational(rootp.ceil, 1)
    2.upto(n){|i|
        r = Rational((rootp * i).floor, i)
        r = Rational([r.numerator, n].min, i)
        left = r if left < r
        r = Rational((rootp * i).ceil, i)
        r = Rational([r.numerator, n].min, i)
        right = r if right > r && r > rootp
    }
    puts "#{right} #{left}"
end