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 isprime(int n) {
  for (int i = 2; i * i <= n; i++) {
    if (!(n % i)) return false;
  }
  return true;
}
int find1(int n) {
  for (; n >= 2; n--) {
    if (isprime(n)) return n;
  }
}
int find2(int n) {
  for (n++;; n++) {
    if (isprime(n)) return n;
  }
}
long long gcd(long long a, long long b) {
  if (a < b) swap(a, b);
  long long tmp = a % b;
  while (tmp) {
    a = b;
    b = tmp;
    tmp = a % b;
  }
  return b;
}
int main() {
  int T, n;
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n);
    long long v = find1(n);
    long long u = find2(n);
    long long a = u * (v - 2) + 2 * (n - v + 1);
    long long b = 2 * u * v;
    long long t = gcd(a, b);
    printf("%I64d/%I64d\n", a / t, b / t);
  }
  return 0;
}
