#include <bits/stdc++.h>
using namespace std;
long long int max(long long int a, long long int b, long long int c);
long long int max(long long int a, long long int b, long long int c) {
  if (a <= c && b <= c) return c;
  if (a >= b && a >= c)
    return a;
  else
    return b;
}
int main() {
  long long int r, g, b, ans1 = 0, ans2 = 0, ans3 = 0;
  cin >> r >> g >> b;
  if (r + g + b < 3) {
    cout << "0";
    return 0;
  }
  if (((r && b) || (b && g) || (g && r)) && (!r || !g || !b) &&
      (r + g + b < 5)) {
    cout << "0";
    return 0;
  }
  ans1 = r / 3 + g / 3 + b / 3;
  if (r && g && b) {
    ans2 += 1;
    ans2 += (r - 1) / 3 + (b - 1) / 3 + (g - 1) / 3;
  }
  if (r > 1 && g > 1 && b > 1) {
    ans3 += 2;
    r -= 2;
    g -= 2;
    b -= 2;
    ans3 += r / 3 + b / 3 + g / 3;
  }
  cout << max(ans1, ans2, ans3);
  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;
long long int max(long long int a, long long int b, long long int c);
long long int max(long long int a, long long int b, long long int c) {
  if (a <= c && b <= c) return c;
  if (a >= b && a >= c)
    return a;
  else
    return b;
}
int main() {
  long long int r, g, b, ans1 = 0, ans2 = 0, ans3 = 0;
  cin >> r >> g >> b;
  if (r + g + b < 3) {
    cout << "0";
    return 0;
  }
  if (((r && b) || (b && g) || (g && r)) && (!r || !g || !b) &&
      (r + g + b < 5)) {
    cout << "0";
    return 0;
  }
  ans1 = r / 3 + g / 3 + b / 3;
  if (r && g && b) {
    ans2 += 1;
    ans2 += (r - 1) / 3 + (b - 1) / 3 + (g - 1) / 3;
  }
  if (r > 1 && g > 1 && b > 1) {
    ans3 += 2;
    r -= 2;
    g -= 2;
    b -= 2;
    ans3 += r / 3 + b / 3 + g / 3;
  }
  cout << max(ans1, ans2, ans3);
  return 0;
}

```