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 gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
bool is_prime(long long x) {
  if (x <= 1) return false;
  for (long long i = 2; i * i <= x; i++)
    if (x % i == 0) return false;
  return true;
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    long long v = n, u = n + 1;
    while (!is_prime(v)) v--;
    while (!is_prime(u)) u++;
    long long num = (v - 2) * u + 2 * (n - v + 1);
    long long den = 2 * v * u;
    long long g = gcd(num, den);
    cout << num / g << "/" << den / g << endl;
  }
  return 0;
}
