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 su(long long x) {
  long long i;
  for (i = 2; i * i <= x; i++)
    if (x % i == 0) return false;
  return true;
}
long long gcd(long long x, long long y) {
  long long temp;
  while (y) {
    temp = y;
    y = x % y;
    x = temp;
  }
  return x;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    long long n, u, v, w, x, y;
    scanf("%lld", &n);
    for (v = n;; v--)
      if (su(v)) break;
    for (u = n + 1;; u++)
      if (su(u)) break;
    x = u * v - 2 * v - 2 * u + 2 * n + 2;
    y = 2 * u * v;
    w = gcd(x, y);
    printf("%lld/%lld\n", x / w, y / w);
  }
  return 0;
}
