#include <bits/stdc++.h>
using namespace std;
long long r, g, b, high = 0;
int main() {
  scanf("%I64d %I64d %I64d", &r, &g, &b);
  high = r / 3 + g / 3 + b / 3 + min(r % 3, min(g % 3, b % 3));
  if (r > 0 && g > 0 && b > 0) {
    if (high < 1 + (r - 1) / 3 + (g - 1) / 3 + (b - 1) / 3 +
                   min((r - 1) % 3, min((g - 1) % 3, (b - 1) % 3))) {
      high = 1 + (r - 1) / 3 + (g - 1) / 3 + (b - 1) / 3 +
             min((r - 1) % 3, min((g - 1) % 3, (b - 1) % 3));
    }
  }
  if (r > 1 && g > 1 && b > 1) {
    if (high < 2 + (r - 2) / 3 + (g - 2) / 3 + (b - 2) / 3 +
                   min((r - 2) % 3, min((g - 2) % 3, (b - 2) % 3))) {
      high = 2 + (r - 2) / 3 + (g - 2) / 3 + (b - 2) / 3 +
             min((r - 2) % 3, min((g - 2) % 3, (b - 2) % 3));
    }
  }
  printf("%d\n", high);
  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;
long long r, g, b, high = 0;
int main() {
  scanf("%I64d %I64d %I64d", &r, &g, &b);
  high = r / 3 + g / 3 + b / 3 + min(r % 3, min(g % 3, b % 3));
  if (r > 0 && g > 0 && b > 0) {
    if (high < 1 + (r - 1) / 3 + (g - 1) / 3 + (b - 1) / 3 +
                   min((r - 1) % 3, min((g - 1) % 3, (b - 1) % 3))) {
      high = 1 + (r - 1) / 3 + (g - 1) / 3 + (b - 1) / 3 +
             min((r - 1) % 3, min((g - 1) % 3, (b - 1) % 3));
    }
  }
  if (r > 1 && g > 1 && b > 1) {
    if (high < 2 + (r - 2) / 3 + (g - 2) / 3 + (b - 2) / 3 +
                   min((r - 2) % 3, min((g - 2) % 3, (b - 2) % 3))) {
      high = 2 + (r - 2) / 3 + (g - 2) / 3 + (b - 2) / 3 +
             min((r - 2) % 3, min((g - 2) % 3, (b - 2) % 3));
    }
  }
  printf("%d\n", high);
  return 0;
}

```