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;
int check(long long num) {
  for (int i = 2; i * i <= num; i++)
    if (num % i == 0) return 0;
  return 1;
}
long long gcd(long long a, long long b) {
  if (a < b) swap(a, b);
  while (b) {
    long long tmp = a % b;
    a = b;
    b = tmp;
  }
  return a;
}
int main() {
  int n, m;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &m);
    long long left = m, right = m + 1;
    while (!check(left)) --left;
    while (!check(right)) ++right;
    long long up = left * right - 2 * right - 2 * left + 2 * m + 2;
    long long down = 2 * left * right;
    long long tmp = gcd(up, down);
    printf("%lld/%lld\n", up / tmp, down / tmp);
  }
}
