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;
using LL = long long int;
LL gcd(LL a, LL b) {
  if (a == 0) return 1;
  if (a < 0) return gcd(-a, b);
  while (b) {
    LL t = a % b;
    a = b;
    b = t;
  }
  return a;
}
template <class T>
bool IsPrime(T n) {
  if (n == 2 || n == 3) return true;
  if (n < 2 || n % 2 == 0) return false;
  if (n < 9) return true;
  if (n % 3 == 0) return false;
  T r = (T)sqrt(n);
  T f = 5;
  while (f <= r) {
    if (n % f == 0) return false;
    if (n % (f + 2) == 0) return false;
    f += 6;
  }
  return true;
}
int n;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    if (n == 2) {
      puts("1/6");
      continue;
    }
    int left = n;
    while (!IsPrime(left)) left--;
    int right = n + 1;
    while (!IsPrime(right)) right++;
    LL up = 1LL * (left - 2) * right + 2LL * (n - left + 1);
    LL down = 2LL * left * right;
    LL g = gcd(up, down);
    up /= g;
    down /= g;
    printf("%lld/%lld\n", up, down);
  }
}
