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;
inline long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
inline bool prime(long long int n) {
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
int main() {
  long long int t;
  cin >> t;
  for (long long int i = 0; i < t; i++) {
    long long int n;
    cin >> n;
    long long int u = n + 1;
    long long int v = n;
    while (!prime(v)) v--;
    while (!prime(u)) u++;
    if (u == n + 1) {
      cout << u - 2 << "/" << 2 * u << endl;
    } else {
      long long int num = (u * (v - 2)) + ((n - v + 1) * 2);
      long long int den = 2 * u * v;
      long long int mcd = gcd(num, den);
      num /= mcd;
      den /= mcd;
      cout << num << "/" << den << endl;
    }
  }
}
