#include <bits/stdc++.h>
using namespace std;
template <typename... T>
inline void input(T &...args) {
  ((cin >> args), ...);
}
template <typename... T>
inline void print(T &&...args) {
  ((cout << args << " "), ...);
  cout << "\n";
}
int solve(int r, int b, int g) {
  return r >= 0 && b >= 0 && g >= 0 ? r / 3 + b / 3 + g / 3 : -1000;
}
int main() {
  cin.tie(nullptr);
  ios_base::sync_with_stdio(false);
  int r, b, g;
  input(r, b, g);
  print(max(max(solve(r, b, g), 1 + solve(r - 1, b - 1, g - 1)),
            2 + solve(r - 2, b - 2, g - 2)));
  return 0;
}

### Prompt

Create a solution in CPP for 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;
template <typename... T>
inline void input(T &...args) {
  ((cin >> args), ...);
}
template <typename... T>
inline void print(T &&...args) {
  ((cout << args << " "), ...);
  cout << "\n";
}
int solve(int r, int b, int g) {
  return r >= 0 && b >= 0 && g >= 0 ? r / 3 + b / 3 + g / 3 : -1000;
}
int main() {
  cin.tie(nullptr);
  ios_base::sync_with_stdio(false);
  int r, b, g;
  input(r, b, g);
  print(max(max(solve(r, b, g), 1 + solve(r - 1, b - 1, g - 1)),
            2 + solve(r - 2, b - 2, g - 2)));
  return 0;
}

```