#include <bits/stdc++.h>
using namespace std;
namespace Sol {
int fd(int r, int g, int b) {
  int ret = 0;
  ret += r / 3;
  ret += g / 3;
  ret += b / 3;
  return ret;
}
void sol() {
  int r, g, b;
  scanf("%d", &r);
  scanf("%d", &g);
  scanf("%d", &b);
  int best = 0;
  int mn = min(min(r, g), b);
  best = mn + fd(r - mn, g - mn, b - mn);
  if (mn - 1 >= 0) {
    best = max(best, mn - 1 + fd(r - mn + 1, g - mn + 1, b - mn + 1));
  }
  if (mn - 2 >= 0) {
    best = max(best, mn - 2 + fd(r - mn + 2, g - mn + 2, b - mn + 2));
  }
  if (mn - 3 >= 0) {
    best = max(best, mn - 3 + fd(r - mn + 3, g - mn + 3, b - mn + 3));
  }
  printf("%d\n", best);
}
}  // namespace Sol
int main() {
  Sol::sol();
  return 0;
}

### Prompt

Your challenge is to write 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;
namespace Sol {
int fd(int r, int g, int b) {
  int ret = 0;
  ret += r / 3;
  ret += g / 3;
  ret += b / 3;
  return ret;
}
void sol() {
  int r, g, b;
  scanf("%d", &r);
  scanf("%d", &g);
  scanf("%d", &b);
  int best = 0;
  int mn = min(min(r, g), b);
  best = mn + fd(r - mn, g - mn, b - mn);
  if (mn - 1 >= 0) {
    best = max(best, mn - 1 + fd(r - mn + 1, g - mn + 1, b - mn + 1));
  }
  if (mn - 2 >= 0) {
    best = max(best, mn - 2 + fd(r - mn + 2, g - mn + 2, b - mn + 2));
  }
  if (mn - 3 >= 0) {
    best = max(best, mn - 3 + fd(r - mn + 3, g - mn + 3, b - mn + 3));
  }
  printf("%d\n", best);
}
}  // namespace Sol
int main() {
  Sol::sol();
  return 0;
}

```