#include <bits/stdc++.h>
using namespace std;
int main() {
  int r, g, b;
  cin >> r >> g >> b;
  int r1, g1, b1, r2, g2, b2;
  r1 = r / 3;
  g1 = g / 3;
  b1 = b / 3;
  r2 = r % 3;
  g2 = g % 3;
  b2 = b % 3;
  int combi = 0;
  while (r2 != 0 && g2 != 0 && b2 != 0) {
    r2 = r2 - 1;
    g2 = g2 - 1;
    b2 = b2 - 1;
    combi++;
  }
  int res = r1 + g1 + b1 + combi;
  if ((r != 0 && g != 0 && b != 0) &&
      ((r2 == 2 && g2 == 2 && b2 == 0) || (r2 == 2 && g2 == 0 && b2 == 2) ||
       (r2 == 0 && g2 == 2 && b2 == 2)))
    res++;
  cout << res << 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;
int main() {
  int r, g, b;
  cin >> r >> g >> b;
  int r1, g1, b1, r2, g2, b2;
  r1 = r / 3;
  g1 = g / 3;
  b1 = b / 3;
  r2 = r % 3;
  g2 = g % 3;
  b2 = b % 3;
  int combi = 0;
  while (r2 != 0 && g2 != 0 && b2 != 0) {
    r2 = r2 - 1;
    g2 = g2 - 1;
    b2 = b2 - 1;
    combi++;
  }
  int res = r1 + g1 + b1 + combi;
  if ((r != 0 && g != 0 && b != 0) &&
      ((r2 == 2 && g2 == 2 && b2 == 0) || (r2 == 2 && g2 == 0 && b2 == 2) ||
       (r2 == 0 && g2 == 2 && b2 == 2)))
    res++;
  cout << res << endl;
}

```