#include <bits/stdc++.h>
using namespace std;
void solve(int test_case) {
  int r, g, b;
  cin >> r >> g >> b;
  int ans = 0;
  for (int mix = 0; mix < 3 && r && g && b; mix++) {
    ans = max(ans, (r - mix) / 3 + (g - mix) / 3 + (b - mix) / 3 + mix);
  }
  if (r == 0 || g == 0 || b == 0) ans = r / 3 + g / 3 + b / 3;
  cout << ans << "\n";
}
int main() {
  int t = 1;
  for (int i = 0; i < t; i++) solve(i);
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
void solve(int test_case) {
  int r, g, b;
  cin >> r >> g >> b;
  int ans = 0;
  for (int mix = 0; mix < 3 && r && g && b; mix++) {
    ans = max(ans, (r - mix) / 3 + (g - mix) / 3 + (b - mix) / 3 + mix);
  }
  if (r == 0 || g == 0 || b == 0) ans = r / 3 + g / 3 + b / 3;
  cout << ans << "\n";
}
int main() {
  int t = 1;
  for (int i = 0; i < t; i++) solve(i);
  return 0;
}

```