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;
long long prevprime(long long x) {
  while (1) {
    bool prime = 1;
    for (int i = 2; i * i <= x; i++) {
      if (x % i == 0) {
        prime = 0;
        break;
      }
    }
    if (prime) return x;
    x--;
  }
}
long long GCD(long long a, long long b) {
  if (b == 0) return a;
  return GCD(b, a % b);
}
long long nextprime(long long x) {
  x++;
  while (1) {
    bool prime = 1;
    for (int i = 2; i * i <= x; i++) {
      if (x % i == 0) {
        prime = 0;
        break;
      }
    }
    if (prime) return x;
    x++;
  }
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long p = prevprime(n), q = nextprime(n);
    long long N, D;
    N = q * p + 2 * (n - q - p + 1);
    D = 2 * p * q;
    long long g = GCD(N, D);
    cout << N / g << "/" << D / g << endl;
  }
}
