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;
bool check(long long int n) {
  if (n == 2 || n == 3) return true;
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
long long int v(long long int n) {
  while (true) {
    if (check(n)) return n;
    n--;
  }
}
long long int u(long long int n) {
  n++;
  while (true) {
    if (check(n)) return n;
    n++;
  }
}
long long int gcd(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
int main() {
  int t;
  long long int n;
  cin >> t;
  while (t--) {
    cin >> n;
    long long int vv = v(n);
    long long int uu = u(n);
    long long int num = uu * (vv - 2) + 2 * (n + 1 - vv);
    long long int den = 2 * uu * vv;
    long long int g = gcd(num, den);
    num = num / g;
    den = den / g;
    cout << num << "/" << den << endl;
  }
}
