#include <bits/stdc++.h>
using namespace std;
int main() {
  int r, g, b;
  cin >> r >> g >> b;
  int kol1 = r / 3;
  int kol2 = g / 3;
  int kol3 = b / 3;
  int ost1 = r % 3;
  int ost2 = g % 3;
  int ost3 = b % 3;
  if (ost1 < ost2) {
    swap(ost1, ost2);
    swap(kol1, kol2);
  }
  if (ost1 < ost3) {
    swap(ost1, ost3);
    swap(kol1, kol3);
  }
  if (ost2 < ost3) {
    swap(ost2, ost3);
    swap(kol2, kol3);
  }
  int ans = 0;
  if (ost1 == 2 && ost2 == 2 && ost3 == 0) {
    if (kol3 > 0) {
      ost3 = 3;
      kol3--;
    }
    ans = kol1 + kol2 + kol3 + min(ost1, min(ost2, ost3));
  } else {
    ans = kol1 + kol2 + kol3 + min(ost1, min(ost2, ost3));
  }
  cout << ans;
  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;
int main() {
  int r, g, b;
  cin >> r >> g >> b;
  int kol1 = r / 3;
  int kol2 = g / 3;
  int kol3 = b / 3;
  int ost1 = r % 3;
  int ost2 = g % 3;
  int ost3 = b % 3;
  if (ost1 < ost2) {
    swap(ost1, ost2);
    swap(kol1, kol2);
  }
  if (ost1 < ost3) {
    swap(ost1, ost3);
    swap(kol1, kol3);
  }
  if (ost2 < ost3) {
    swap(ost2, ost3);
    swap(kol2, kol3);
  }
  int ans = 0;
  if (ost1 == 2 && ost2 == 2 && ost3 == 0) {
    if (kol3 > 0) {
      ost3 = 3;
      kol3--;
    }
    ans = kol1 + kol2 + kol3 + min(ost1, min(ost2, ost3));
  } else {
    ans = kol1 + kol2 + kol3 + min(ost1, min(ost2, ost3));
  }
  cout << ans;
  return 0;
}

```