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;
const double PI = 2.0 * acos(0.0);
const double EPS = 1e-6;
bool is_prime(int x) {
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) return false;
  return true;
}
long long gcd(long long a, long long b) {
  for (long long tmp = a % b; tmp != 0; tmp = a % b) {
    a = b;
    b = tmp;
  }
  return b;
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    int n;
    scanf("%d", &n);
    int p = n, q = n + 1;
    for (; !is_prime(p); p--)
      ;
    for (; !is_prime(q); q++)
      ;
    long long upper = 2LL * (1LL + n - p - q) + (long long)p * q;
    long long lower = 2LL * p * q;
    long long g = gcd(upper, lower);
    printf("%I64d/%I64d\n", upper / g, lower / g);
  }
  return 0;
}
