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;
const int MOD = 1e9 + 7;
bool isprim(long long int n) {
  long long int i;
  for (i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    long long int p, q;
    p = q = n + 1;
    while (isprim(p) == false) p--;
    while (isprim(q) == false) q++;
    long long int t1, t2;
    t1 = p * q - 2 * q + 2 * (n - p + 1);
    t2 = 2 * p * q;
    long long int temp = gcd(t1, t2);
    cout << t1 / temp << "/" << t2 / temp << endl;
  }
}
