#include <bits/stdc++.h>
using namespace std;
long long int ans(long long int r, long long int g, long long int b) {
  if (r == g && r == b && g == b) {
    return r;
  }
  long long int R(r % 3), G(g % 3), B(b % 3);
  long long int ans = (r / 3) + (b / 3) + (g / 3) + min(B, min(R, G));
  if ((R == 2 && B == 2 && G == 0 && g != 0) ||
      (G == 2 && B == 2 && R == 0 && r != 0) ||
      (G == 2 && R == 2 && B == 0 && b != 0)) {
    ans++;
  }
  return ans;
}
int main() {
  long long int r, g, b;
  cin >> r >> g >> b;
  cout << ans(r, g, b) << endl;
}

### Prompt

Please formulate a cpp solution 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;
long long int ans(long long int r, long long int g, long long int b) {
  if (r == g && r == b && g == b) {
    return r;
  }
  long long int R(r % 3), G(g % 3), B(b % 3);
  long long int ans = (r / 3) + (b / 3) + (g / 3) + min(B, min(R, G));
  if ((R == 2 && B == 2 && G == 0 && g != 0) ||
      (G == 2 && B == 2 && R == 0 && r != 0) ||
      (G == 2 && R == 2 && B == 0 && b != 0)) {
    ans++;
  }
  return ans;
}
int main() {
  long long int r, g, b;
  cin >> r >> g >> b;
  cout << ans(r, g, b) << endl;
}

```