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 prime(long long x) {
  if (x % 2 == 0) return false;
  for (long long j = 3; j * j <= x; j += 2)
    if (x % j == 0) return false;
  return true;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  a %= b;
  return gcd(b, a);
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    long long n;
    cin >> n;
    long long v = n, u = n + 1;
    while (!prime(v)) v--;
    while (!prime(u)) u++;
    long long low = v * u * 2;
    long long up = v * u - 2 * v - 2 * (u - n - 1);
    long long g = gcd(low, up);
    up /= g, low /= g;
    printf("%I64d/", up);
    printf("%I64d\n", low);
  }
  return 0;
}
