#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int r, g, b, res = 0, flag = 0, sum;
  cin >> r;
  cin >> g;
  cin >> b;
  sum = r + g + b;
  if (r == 0 || g == 0 || b == 0) {
    cout << r / 3 + g / 3 + b / 3;
    return 0;
  } else if (r % 3 != g % 3 && r % 3 != b % 3 && b % 3 != g % 3) {
    if (r < b && r < g) {
      cout << r + (g - r) / 3 + (b - r) / 3;
      return 0;
    } else if (b < g) {
      cout << b + (g - b) / 3 + (r - b) / 3;
      return 0;
    } else {
      cout << g + (r - g) / 3 + (b - g) / 3;
      return 0;
    }
  } else {
    cout << sum / 3;
    return 0;
  }
  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() {
  long long int r, g, b, res = 0, flag = 0, sum;
  cin >> r;
  cin >> g;
  cin >> b;
  sum = r + g + b;
  if (r == 0 || g == 0 || b == 0) {
    cout << r / 3 + g / 3 + b / 3;
    return 0;
  } else if (r % 3 != g % 3 && r % 3 != b % 3 && b % 3 != g % 3) {
    if (r < b && r < g) {
      cout << r + (g - r) / 3 + (b - r) / 3;
      return 0;
    } else if (b < g) {
      cout << b + (g - b) / 3 + (r - b) / 3;
      return 0;
    } else {
      cout << g + (r - g) / 3 + (b - g) / 3;
      return 0;
    }
  } else {
    cout << sum / 3;
    return 0;
  }
  return 0;
}

```