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 m, long long n) {
  if (m % n == 0) return n;
  return gcd(n, m % n);
}
bool isPrime(long long K) {
  for (int i = 2; i * i <= K; i++)
    if (K % i == 0) return false;
  return true;
}
int main() {
  int T;
  cin >> T;
  for (int i = 0; i < T; i++) {
    int N;
    cin >> N;
    long long p;
    for (p = N; p >= 2; p--)
      if (isPrime(p)) break;
    long long q;
    for (q = N + 1; q <= 1000000007; q++)
      if (isPrime(q)) break;
    long long num = p * q - 2 * q + 2 * (N - p + 1);
    long long dem = 2 * p * q;
    long long d = gcd(num, dem);
    cout << num / d << "/" << dem / d << "\n";
  }
  return 0;
}
