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 Cas, n, l, r, ansm, ansn, Gcd;
long long Getgcd(long long x, long long y) {
  for (long long t; y; t = x % y, x = y, y = t)
    ;
  return x;
}
int Check(long long x) {
  for (long long i = 2; i * i <= x; ++i)
    if (x % i == 0) return 0;
  return 1;
}
int main() {
  scanf("%I64d", &Cas);
  while (Cas--) {
    scanf("%d", &n);
    for (l = n; !Check(l); --l)
      ;
    for (r = n + 1; !Check(r); ++r)
      ;
    ansm = 2 * l * r;
    ansn = (l - 2) * r + (n - l + 1) * 2;
    Gcd = Getgcd(ansn, ansm);
    printf("%I64d/%I64d\n", ansn / Gcd, ansm / Gcd);
  }
}
