#include <bits/stdc++.h>
using namespace std;
const double PI = 3.141592653589793238;
template <typename T>
std::ostream& operator<<(std::ostream& os, const std::vector<T>& vector) {
  for (size_t i = 0; i < vector.size(); ++i) {
    os << vector[i] << " ";
  }
  return os;
}
long long r, g, b, ans = 0;
int main() {
  std::ios_base::sync_with_stdio(false);
  cin >> r >> g >> b;
  for (int i = 0; i < 3; ++i) {
    long long tmp = 0, r_t = r, g_t = g, b_t = b;
    for (int j = 1; j <= i; ++j) {
      if (r_t == 0 || g_t == 0 || b_t == 0) {
        break;
      }
      --r_t, --g_t, --b_t;
      ++tmp;
    }
    tmp += r_t / 3 + g_t / 3 + b_t / 3;
    ans = max(ans, tmp);
  }
  cout << ans << "\n";
  return 0;
}

### Prompt

Please create a solution in Cpp to the following problem:

Fox Ciel has some flowers: r red flowers, g green flowers and b blue flowers. She wants to use these flowers to make several bouquets. There are 4 types of bouquets:

  * To make a "red bouquet", it needs 3 red flowers. 
  * To make a "green bouquet", it needs 3 green flowers. 
  * To make a "blue bouquet", it needs 3 blue flowers. 
  * To make a "mixing bouquet", it needs 1 red, 1 green and 1 blue flower. 



Help Fox Ciel to find the maximal number of bouquets she can make.

Input

The first line contains three integers r, g and b (0 ≤ r, g, b ≤ 109) — the number of red, green and blue flowers.

Output

Print the maximal number of bouquets Fox Ciel can make.

Examples

Input

3 6 9


Output

6


Input

4 4 4


Output

4


Input

0 0 0


Output

0

Note

In test case 1, we can make 1 red bouquet, 2 green bouquets and 3 blue bouquets.

In test case 2, we can make 1 red, 1 green, 1 blue and 1 mixing bouquet.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const double PI = 3.141592653589793238;
template <typename T>
std::ostream& operator<<(std::ostream& os, const std::vector<T>& vector) {
  for (size_t i = 0; i < vector.size(); ++i) {
    os << vector[i] << " ";
  }
  return os;
}
long long r, g, b, ans = 0;
int main() {
  std::ios_base::sync_with_stdio(false);
  cin >> r >> g >> b;
  for (int i = 0; i < 3; ++i) {
    long long tmp = 0, r_t = r, g_t = g, b_t = b;
    for (int j = 1; j <= i; ++j) {
      if (r_t == 0 || g_t == 0 || b_t == 0) {
        break;
      }
      --r_t, --g_t, --b_t;
      ++tmp;
    }
    tmp += r_t / 3 + g_t / 3 + b_t / 3;
    ans = max(ans, tmp);
  }
  cout << ans << "\n";
  return 0;
}

```