#include <bits/stdc++.h>
using namespace std;
int main() {
  long long x[5];
  for (long long i = 0; i < 3; i++) {
    cin >> x[i];
  }
  sort(x, x + 3);
  long long d = x[0];
  x[1] = x[1] - x[0];
  x[2] = x[2] - x[0];
  if (((x[1] % 3 == 2) && x[2] % 3 == 2) && (x[0] > 0)) {
    d++;
  }
  d += x[1] / 3;
  d += x[2] / 3;
  cout << d << endl;
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 x[5];
  for (long long i = 0; i < 3; i++) {
    cin >> x[i];
  }
  sort(x, x + 3);
  long long d = x[0];
  x[1] = x[1] - x[0];
  x[2] = x[2] - x[0];
  if (((x[1] % 3 == 2) && x[2] % 3 == 2) && (x[0] > 0)) {
    d++;
  }
  d += x[1] / 3;
  d += x[2] / 3;
  cout << d << endl;
  return 0;
}

```