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>
long long gcd(long long x, long long y) { return x % y ? gcd(y, x % y) : y; }
bool is_prime(long long n) {
  for (long long i = 2; i * i <= n; ++i)
    if (!(n % i)) return false;
  return n != 1;
}
int main() {
  int t;
  long long n, p0, p1, a, b, c;
  scanf("%d", &t);
  while (t--) {
    scanf("%I64d", &n);
    for (p0 = n; !is_prime(p0); --p0)
      ;
    for (p1 = n + 1; !is_prime(p1); ++p1)
      ;
    a = p0 * p1 + 2 * (n + 1 - p0 - p1);
    b = 2 * p0 * p1;
    c = gcd(a, b);
    printf("%I64d/%I64d\n", a / c, b / c);
  }
  return 0;
}
