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;
inline bool pd(long long x) {
  for (long long i = 2; i * i <= x; ++i) {
    if (x % i == 0) return 0;
  }
  return 1;
}
long long findu(long long x) {
  while (!pd(x)) --x;
  return x;
}
long long findv(long long x) {
  while (!pd(x)) ++x;
  return x;
}
inline long long gcd(long long x, long long y) {
  if (x < y) swap(x, y);
  return y ? gcd(y, x % y) : x;
}
long long t, n;
signed main() {
  ios::sync_with_stdio(false);
  cin >> t;
  long long l, r, x, y, qwq;
  while (t--) {
    cin >> n;
    l = findu(n);
    r = findv(n + 1);
    x = (l - 2) * r + 2 * (n - l + 1);
    y = 2 * l * r;
    qwq = gcd(x, y);
    cout << x / qwq << "/" << y / qwq << "\n";
  }
  return 0;
}
