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