#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long r, g, b;
  cin >> r >> g >> b;
  long long clr[3] = {r, g, b};
  sort(clr, clr + 3);
  long long ans = 0;
  if (r >= 2 && g >= 2 && b >= 2)
    ans = max(r / 3 + g / 3 + b / 3,
              max(1 + (r - 1) / 3 + (g - 1) / 3 + (b - 1) / 3,
                  2 + (r - 2) / 3 + (g - 2) / 3 + (b - 2) / 3));
  else {
    ans = clr[0] / 3;
    ans *= 3;
    long long rm = clr[0] % 3;
    for (long long i = 0; i < 3; i++) {
      clr[i] -= ans;
    }
    if (rm == 2) {
      ans += 2;
      for (long long i = 0; i < 3; i++) {
        clr[i] -= 2;
      }
    } else if (rm == 1) {
      ans += 1;
      for (long long i = 0; i < 3; i++) {
        clr[i] -= 1;
      }
    }
    ans += clr[1] / 3 + clr[2] / 3;
  }
  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() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long r, g, b;
  cin >> r >> g >> b;
  long long clr[3] = {r, g, b};
  sort(clr, clr + 3);
  long long ans = 0;
  if (r >= 2 && g >= 2 && b >= 2)
    ans = max(r / 3 + g / 3 + b / 3,
              max(1 + (r - 1) / 3 + (g - 1) / 3 + (b - 1) / 3,
                  2 + (r - 2) / 3 + (g - 2) / 3 + (b - 2) / 3));
  else {
    ans = clr[0] / 3;
    ans *= 3;
    long long rm = clr[0] % 3;
    for (long long i = 0; i < 3; i++) {
      clr[i] -= ans;
    }
    if (rm == 2) {
      ans += 2;
      for (long long i = 0; i < 3; i++) {
        clr[i] -= 2;
      }
    } else if (rm == 1) {
      ans += 1;
      for (long long i = 0; i < 3; i++) {
        clr[i] -= 1;
      }
    }
    ans += clr[1] / 3 + clr[2] / 3;
  }
  cout << ans;
  return 0;
}

```