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>
int isPrime(long long n) {
  long long i;
  if (n == 1) return 0;
  for (i = 2; i <= sqrt(n); i++)
    if (n % i == 0) return 0;
  return 1;
}
long long gcd(long long a, long long b) {
  long long t;
  if (a < b) {
    t = a;
    a = b;
    b = t;
  }
  while (b != 0) {
    t = a % b;
    a = b;
    b = t;
  }
  return a;
}
int main() {
  long long t, n, i, t1, t2, p, q, g;
  scanf("%I64d", &t);
  while (t--) {
    scanf("%I64d", &n);
    p = n;
    while (isPrime(p) == 0) p--;
    q = n + 1;
    while (isPrime(q) == 0) q++;
    t1 = p * q - 2 * q + 2 * (n - p + 1);
    t2 = 2 * p * q;
    g = gcd(t1, t2);
    printf("%I64d/%I64d\n", t1 / g, t2 / g);
  }
  return 0;
}
