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 prime(long long 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) { return a % b ? gcd(b, a % b) : b; }
int main(void) {
  int T;
  scanf("%d", &T);
  while (T--) {
    long long n;
    scanf("%I64d", &n);
    long long pk = n, pkk = n + 1;
    while (!prime(pk)) pk--;
    while (!prime(pkk)) pkk++;
    long long a = pk * pkk - 2 * pkk + 2 * (n - pk + 1);
    long long b = 2 * pk * pkk;
    long long g = gcd(a, b);
    printf("%I64d/%I64d\n", a / g, b / g);
  }
  return 0;
}
