#include <bits/stdc++.h>
using namespace std;
long long bsearch(int x, int y, int b) {
  long long low = 1;
  long long high = 2e10;
  while (low <= high) {
    long long mid = (low + high) / 2;
    long long val = ((long long)x * b) - (mid * y);
    if (val == 0) return mid;
    if (val > 0)
      low = mid + 1;
    else
      high = mid - 1;
  }
  return (low + high) / 2;
}
int main() {
  int x, y, n;
  cin >> x >> y >> n;
  long long best_diff = 1e9;
  long long best_diff_den = 1;
  long long resa, resb;
  for (int b = 1; b <= n; b++) {
    long long a = bsearch(x, y, b);
    if (b == 49351) {
      a = bsearch(x, y, b);
    }
    for (int offset = -1; offset <= 1; offset++) {
      long long a1 = a + offset;
      long long diff = ((long long)x * (long long)b - a1 * (long long)y);
      long long diff_den = (long long)y * b;
      diff = abs(diff);
      if (best_diff * diff_den - best_diff_den * diff > 0) {
        best_diff = diff;
        best_diff_den = diff_den;
        resa = a1;
        resb = b;
      }
    }
  }
  cout << resa << "/" << resb << endl;
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

You are given three positive integers x, y, n. Your task is to find the nearest fraction to fraction <image> whose denominator is no more than n. 

Formally, you should find such pair of integers a, b (1 ≤ b ≤ n; 0 ≤ a) that the value <image> is as minimal as possible.

If there are multiple "nearest" fractions, choose the one with the minimum denominator. If there are multiple "nearest" fractions with the minimum denominator, choose the one with the minimum numerator.

Input

A single line contains three integers x, y, n (1 ≤ x, y, n ≤ 105).

Output

Print the required fraction in the format "a/b" (without quotes).

Examples

Input

3 7 6


Output

2/5


Input

7 2 4


Output

7/2

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long bsearch(int x, int y, int b) {
  long long low = 1;
  long long high = 2e10;
  while (low <= high) {
    long long mid = (low + high) / 2;
    long long val = ((long long)x * b) - (mid * y);
    if (val == 0) return mid;
    if (val > 0)
      low = mid + 1;
    else
      high = mid - 1;
  }
  return (low + high) / 2;
}
int main() {
  int x, y, n;
  cin >> x >> y >> n;
  long long best_diff = 1e9;
  long long best_diff_den = 1;
  long long resa, resb;
  for (int b = 1; b <= n; b++) {
    long long a = bsearch(x, y, b);
    if (b == 49351) {
      a = bsearch(x, y, b);
    }
    for (int offset = -1; offset <= 1; offset++) {
      long long a1 = a + offset;
      long long diff = ((long long)x * (long long)b - a1 * (long long)y);
      long long diff_den = (long long)y * b;
      diff = abs(diff);
      if (best_diff * diff_den - best_diff_den * diff > 0) {
        best_diff = diff;
        best_diff_den = diff_den;
        resa = a1;
        resb = b;
      }
    }
  }
  cout << resa << "/" << resb << endl;
  return 0;
}

```