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;
const int S = 10;
long long mult_mod(long long a, long long b, long long c) {
  a %= c;
  b %= c;
  long long ret = 0;
  while (b) {
    if (b & 1) {
      ret += a;
      ret %= c;
    }
    a <<= 1;
    if (a >= c) a %= c;
    b >>= 1;
  }
  return ret;
}
long long pow_mod(long long x, long long n, long long mod) {
  if (n == 1) return x % mod;
  x %= mod;
  long long tmp = x;
  long long ret = 1;
  while (n) {
    if (n & 1) ret = mult_mod(ret, tmp, mod);
    tmp = mult_mod(tmp, tmp, mod);
    n >>= 1;
  }
  return ret;
}
bool check(long long a, long long n, long long x, long long t) {
  long long ret = pow_mod(a, x, n);
  long long last = ret;
  for (int i = 1; i <= t; i++) {
    ret = mult_mod(ret, ret, n);
    if (ret == 1 && last != 1 && last != n - 1) return true;
    last = ret;
  }
  if (ret != 1) return true;
  return false;
}
bool Miller_Rabin(long long n) {
  if (n < 2) return false;
  if (n == 2) return true;
  if ((n & 1) == 0) return false;
  long long x = n - 1;
  long long t = 0;
  while ((x & 1) == 0) {
    x >>= 1;
    t++;
  }
  for (int i = 0; i < S; i++) {
    long long a = rand() % (n - 1) + 1;
    if (check(a, n, x, t)) return false;
  }
  return true;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    long long n;
    cin >> n;
    long long m, m1;
    for (m = n + 1;; m++) {
      if (Miller_Rabin(m)) break;
    }
    for (m1 = n;; m1--) {
      if (Miller_Rabin(m1)) break;
    }
    long long x = m * m1 - 2 * (m1 + m - n - 1);
    long long y = 2 * m1 * m;
    if (x % 2 == 0) {
      x /= 2;
      y /= 2;
    }
    if (x % m1 == 0) {
      x /= m1;
      y /= m1;
    }
    if (x % m == 0) {
      x /= m;
      y /= m;
    }
    cout << x << '/' << y << endl;
  }
  return 0;
}
