In this physics problem, what we are concerned about are only resistors. If you are poor at physics, do not worry, since solving this problem does not require you to have advanced abilities in physics.

Resistors are said to be connected together in parallel when both of their terminals are respectively connected to each terminal of the other resistors.

We have the following parallel resistor equation for k resistors with resistances R1, R2, ..., Rk in parallel and their combined resistance R:

Now you have n resistors, the i-th of which has a resistance of ri ohms with the equation

You also have n selections, the i-th of which is a set of resistors Si such that

Please find a selection in which the resistors form a parallel resistor with the minimum resistance and output the reduced fraction  of its resistance.

Input

The input contains several test cases, and the first line contains a positive integer T indicating the number of test cases which is up to 100.

For each test case, the only one line contains an integer n, where 1 ≤ n ≤ 10100.

Output

For each test case, output a line containing a reduced fraction of the form p/q indicating the minimum possible resistance, where p and q should be positive numbers that are coprime.

Example

Input

3
10
100
1000

Output

1/2
5/12
35/96

题意：选一个小于等于n（10^100）的无平方因子的数，它的电阻阻值就是它的所有因子的倒数和的倒数，求这个最小的阻值。

思路：队友(lzw巨佬)猜了一个结论：小于n的连续的素数之积就是满足条件的，它的阻值就是因子倒数和的倒数，因为我们要尽可能使R1,R2……Rk最小，才能使他们的和最大，最后使答案最小。因为是数据特别大，所以我用的java。

import java.lang.reflect.Array;
import java.math.*;
import java.util.*;
public class Main {
	public static int[] getPrimes(int n) {
        int[] a = new int[n];
        for(int i = 2; i < n; i ++) {
            a[i] = i;
        }
        for(int i = 2; i < n; i ++) {
            if (a[i] != 0) {
                for(int j = i * 2; j < n; j = j + i) {
                    a[j] = 0;
                }
            }
        }
        int count = 0;
        for(int i = 2; i < n; i++) {
            if (a[i] != 0) {
                count ++;
            }
        }
        if (count > 0) {
            int[] primes  = new int[count];
            int j = 0;
            for (int i = 2; i < n; i ++) {
                if(a[i] != 0) {
                    primes[j] = a[i];
                    j ++;
                }
            }
            return primes;
        }
        return null;
    }
	public static BigInteger gcd(BigInteger a,BigInteger b) {
        if(b.compareTo(BigInteger.ZERO)==0)
            return a;
        else
            return gcd(b,a.remainder(b));
    }
	public static void main(String[] args) {
		int t;
		Scanner cin=new Scanner(System.in);
		int []prime =getPrimes(5000);
	/*	for(int i=0;i<Array.getLength(prime);i++)
		{
			System.out.println(prime[i]);
		}*/
		t=cin.nextInt();
		while(t--!=0)
		{
			BigInteger n=cin.nextBigInteger();
			int i;
			BigInteger sum=BigInteger.ONE,s=BigInteger.ONE;
			
			for( i=0;;i++)
			{
				if(sum.multiply(BigInteger.valueOf(prime[i])).compareTo(n)<=0)
				{
					sum=sum.multiply(BigInteger.valueOf(prime[i]));
					s=s.multiply(BigInteger.valueOf(prime[i]+1));
				}
				else 
					break;
			}
			BigInteger g=gcd(sum,s);
			s=s.divide(g);
			sum=sum.divide(g);
			System.out.println(sum+"/"+s);
		}
	}
}