Let's assume that 

  * v(n) is the largest prime number, that does not exceed n;
  * u(n) is the smallest prime number strictly greater than n. 



Find <image>.

Input

The first line contains integer t (1 ≤ t ≤ 500) — the number of testscases. 

Each of the following t lines of the input contains integer n (2 ≤ n ≤ 109).

Output

Print t lines: the i-th of them must contain the answer to the i-th test as an irreducible fraction "p/q", where p, q are integers, q > 0.

Examples

Input

2
2
3


Output

1/6
7/30
#include <bits/stdc++.h>
using namespace std;
int t, n;
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
int isPrime(long long a) {
  for (long long i = 2; i * i <= a; ++i)
    if (a % i == 0) return 0;
  return 1;
}
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    long long q = n, r = n + 1;
    while (!isPrime(q)) q--;
    while (!isPrime(r)) r++;
    long long nom = r * ((long long)q - 2) + 2L * ((long long)n - q + 1);
    long long den = 2L * q * r;
    long long g = gcd(nom, den);
    nom /= g;
    den /= g;
    printf("%I64d/%I64d\n", nom, den);
  }
  return 0;
}
