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>
bool is_prime(long long x) {
  if (x < 2) return false;
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) return false;
  }
  return true;
}
long long gcd(long long a, long long b) { return a == 0 ? b : gcd(b % a, a); }
int main() {
  int t;
  scanf("%d", &t);
  long long n;
  while (t--) {
    scanf("%lld", &n);
    long long L = n, R = n + 1;
    while (!is_prime(L)) L--;
    while (!is_prime(R)) R++;
    long long num = L * R - 2 * R + 2 * (n - L + 1);
    long long den = 2 * L * R;
    long long G = gcd(num, den);
    printf("%lld/%lld\n", num / G, den / G);
  }
  return 0;
}
