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;
bool isp(long long x) {
  for (long long i = 2; i * i <= x; i++)
    if (x % i == 0) return 0;
  return !(x < 2);
}
long long gcd(long long x, long long y) {
  if (y != 0)
    return gcd(y, x % y);
  else
    return x;
}
int main() {
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long s1, s2, dp = n, up = n + 1;
    while (!isp(dp)) dp--;
    while (!isp(up)) up++;
    s1 = dp * up - 2 * (dp + up - n - 1);
    s2 = dp * up * 2;
    long long g = gcd(s1, s2);
    cout << s1 / g << "/" << s2 / g << endl;
  }
  return 0;
}
