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;
const int N = 500010;
pair<int, int> a[300300];
long long int gcd(long long int a, long long int b) {
  if (!a)
    return b;
  else
    return gcd(b % a, a);
}
bool prime(long long n) {
  for (long long i = 2; i * i <= n; i++)
    if (!(n % i)) return false;
  return true;
}
int main() {
  int t;
  cin >> t;
  long long n;
  long long u, v;
  unsigned long long int tuso, mauso, ucln;
  while (t--) {
    cin >> n;
    if (prime(n + 1)) {
      tuso = n + 1 - 2;
      mauso = (n + 1) * 2;
      ucln = gcd(tuso, mauso);
      tuso /= ucln;
      mauso /= ucln;
      cout << tuso << "/" << mauso << endl;
    } else {
      u = n;
      v = n + 1;
      while (!prime(u)) {
        u--;
      }
      while (!prime(v)) {
        v++;
      }
      tuso = u * v + 2 * (n + 1 - u - v);
      mauso = 2 * u * v;
      ucln = gcd(tuso, mauso);
      tuso /= ucln;
      mauso /= ucln;
      cout << tuso << "/" << mauso << endl;
    }
  }
  return 0;
}
