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;
long long T, n, v, u;
bool prime(long long n) {
  for (long long i = 2; i * i <= n; ++i)
    if (n % i == 0) return 0;
  return 1;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
struct fraction {
  long long p, q;
  fraction(long long _p = 1, long long _q = 1) {
    p = _p;
    q = _q;
  }
  friend fraction operator-(fraction a, fraction b) {
    fraction c;
    long long tmp = gcd(a.q, b.q);
    c.p = b.q / tmp * a.p - a.q / tmp * b.p;
    c.q = a.q / tmp * b.q;
    tmp = gcd(c.p, c.q);
    c.p /= tmp;
    c.q /= tmp;
    return c;
  }
} a, b;
int main(void) {
  std::ios::sync_with_stdio(false);
  cin >> T;
  while (T--) {
    cin >> n;
    v = n;
    u = n + 1;
    while (!prime(v)) v--;
    while (!prime(u)) u++;
    a = fraction(u - 2, 2 * u);
    b = fraction(u - 1 - n, v * u);
    a = a - b;
    cout << a.p << "/" << a.q << endl;
  }
}
