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;
bool judage(long long x) {
  for (int i = 2; i * i <= x; i++) {
    if (x % i == 0) return false;
  }
  return true;
}
long long gcd(long long x, long long y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
int main() {
  int t;
  long long n, u, v;
  cin >> t;
  while (t--) {
    cin >> n;
    v = n;
    u = n + 1;
    while (!judage(v)) {
      v--;
    }
    while (!judage(u)) {
      u++;
    }
    long long p = u * v + 2 * (n - v - u + 1);
    long long q = 2 * v * u;
    long long mod = gcd(p, q);
    cout << p / mod << "/" << q / mod << endl;
  }
  return 0;
}
