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;
const int NICO = 10000 + 10;
int T, n;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
bool isprime(int x) {
  if (x == 2) return 1;
  for (int i = 2; i * i <= x; i++) {
    if (x % i == 0) return 0;
  }
  return 1;
}
int main() {
  cin >> T;
  while (T--) {
    scanf("%d", &n);
    int tmp = n;
    while (!isprime(tmp)) {
      tmp--;
    }
    long long x = tmp;
    tmp = n + 1;
    while (!isprime(tmp)) {
      tmp++;
    }
    long long y = tmp;
    long long p = x * y - 2 * y + 2 * (n - x + 1);
    long long q = 2 * x * y;
    long long t = gcd(p, q);
    p /= t, q /= t;
    printf("%lld/%lld\n", p, q);
  }
}
