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>
using namespace std;
bool isprime(int x) {
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) return false;
  return true;
}
long long gao(long long a, long long b) {
  if (a < b) swap(a, b);
  if (b == 0) return a;
  return gao(b, a % b);
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    long long n, x, y;
    cin >> n;
    x = y = n;
    while (1) {
      if (isprime(x)) break;
      x--;
    }
    y++;
    while (1) {
      if (isprime(y)) break;
      y++;
    }
    long long q = 2 * x * y, p = x * y - 2 * x - 2 * y + 2 * n + 2;
    long long gcd = gao(q, p);
    printf("%I64d/%I64d\n", p / gcd, q / gcd);
  }
  return 0;
}
