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 isprime(long long n) {
  bool b = true;
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t, n, p1, p2;
  cin >> t;
  for (int i = 0; i < t; i++) {
    cin >> n;
    p1 = n + 1;
    p2 = n;
    while (!isprime(p1)) p1++;
    while (!isprime(p2)) p2--;
    long long op1 = p2 * p1 - 2 * p2 + 2 * n - 2 * (p1 - 1);
    long long op2 = 2 * p1 * p2;
    long long g = gcd(max(op1, op2), min(op1, op2));
    cout << op1 / g << '/' << op2 / g << endl;
  }
}
