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;
int t;
bool isprime(long long x) {
  if (x == 2) return true;
  if (x % 2 == 0) return false;
  long long k = sqrt(x);
  for (long long i = 2; i <= k; i++)
    if (x % i == 0) return false;
  return true;
}
long long under(long long x) {
  for (long long i = x; i >= 2; i--)
    if (isprime(i)) return i;
}
long long top(long long x) {
  for (long long i = x + 1; i; i++)
    if (isprime(i)) return i;
}
long long gcd(long long x, long long y) {
  if (x < y) swap(x, y);
  if (x % y == 0)
    return y;
  else
    return (gcd(y, x % y));
}
int main() {
  scanf("%d", &t);
  int n;
  for (int i = 1; i <= t; i++) {
    scanf("%d", &n);
    long long v = under(n);
    long long u = top(n);
    long long p = v * u - 2 * v - 2 * u + 2 + 2 * n;
    long long q = 2 * v * u;
    long long k = gcd(p, q);
    printf("%I64d/%I64d\n", p / k, q / k);
  }
  return 0;
}
