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 prime(long long n) {
  for (long long i = 2; i <= sqrt(n); i++)
    if (n % i == 0) return 0;
  return 1;
}
long long v(long long x) {
  for (; x > 2; x--)
    if (prime(x)) return x;
}
long long u(long long x) {
  for (x++;; x++)
    if (prime(x)) return x;
}
long long gcd(long long x, long long y) { return y == 0 ? x : gcd(y, x % y); }
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    int n;
    scanf("%d", &n);
    long long a, b, z, w;
    a = u(n);
    z = v(n);
    b = a * (z - 2) + 2 * (n - z) + 2;
    w = 2 * a * z;
    printf("%lld/%lld\n", b / gcd(b, w), w / gcd(b, w));
  }
  return 0;
}
