#include <bits/stdc++.h>
using namespace std;
const int inf = (int)1e9;
const double eps = 1e-9;
const double pi = acos(-1.0);
int main() {
  vector<int> a(3);
  for (int i = 0; i < 3; i++) scanf("%d", &a[i]);
  sort(a.begin(), a.end());
  int mini = a[0] + (a[1] - a[0]) / 3 + (a[2] - a[0]) / 3;
  if (a[0] > 0) {
    a[0]--;
    mini = max(mini, a[0] + (a[1] - a[0]) / 3 + (a[2] - a[0]) / 3);
    if (a[0] > 0) {
      a[0]--;
      mini = max(mini, a[0] + (a[1] - a[0]) / 3 + (a[2] - a[0]) / 3);
    }
  }
  cout << mini << endl;
  return 0;
}

### Prompt

Your task is to create 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;
const int inf = (int)1e9;
const double eps = 1e-9;
const double pi = acos(-1.0);
int main() {
  vector<int> a(3);
  for (int i = 0; i < 3; i++) scanf("%d", &a[i]);
  sort(a.begin(), a.end());
  int mini = a[0] + (a[1] - a[0]) / 3 + (a[2] - a[0]) / 3;
  if (a[0] > 0) {
    a[0]--;
    mini = max(mini, a[0] + (a[1] - a[0]) / 3 + (a[2] - a[0]) / 3);
    if (a[0] > 0) {
      a[0]--;
      mini = max(mini, a[0] + (a[1] - a[0]) / 3 + (a[2] - a[0]) / 3);
    }
  }
  cout << mini << endl;
  return 0;
}

```