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 gcd(long long A, long long B) {
  while (B != 0) {
    A %= B;
    swap(A, B);
  }
  return A;
}
bool prime(int n) {
  if (n == 1) return false;
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
int prev_prime(int n) {
  if (n == 2) return n;
  if (n % 2 == 0) n--;
  while (!prime(n)) n -= 2;
  return n;
}
int next_prime(int n) {
  n++;
  if (n % 2 == 0) n++;
  while (!prime(n)) n += 2;
  return n;
}
void solve(int n, long long& num, long long& denum) {
  long long lo = prev_prime(n), hi = next_prime(n);
  num = lo * hi - 2 * hi + 2 * (n - lo + 1);
  denum = 2 * lo * hi;
  long long g = gcd(num, denum);
  num /= g;
  denum /= g;
}
int main() {
  int t;
  cin >> t;
  long long num, denum;
  for (int i = 0; i < t; i++) {
    int n;
    cin >> n;
    solve(n, num, denum);
    cout << num << '/' << denum << '\n';
  }
  return 0;
}
