#include <bits/stdc++.h>
using namespace std;
const int OO = (int)2e9;
const double eps = 1e-9;
int main() {
  int r, g, b, ret, mn = 0, tr, tg, tb;
  cin >> r >> g >> b;
  if (!r || !g || !b)
    cout << r / 3 + g / 3 + b / 3 << endl;
  else {
    tr = r, tg = g, tb = b;
    mn = min(r, min(g, b));
    r -= mn, g -= mn, b -= mn;
    mn += r / 3 + g / 3 + b / 3;
    ret = mn, mn = 0;
    mn = min(tr, min(tg, tb)) - 1;
    tr -= mn, tg -= mn, tb -= mn;
    mn += tr / 3 + tg / 3 + tb / 3;
    cout << max(mn, ret) << endl;
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 OO = (int)2e9;
const double eps = 1e-9;
int main() {
  int r, g, b, ret, mn = 0, tr, tg, tb;
  cin >> r >> g >> b;
  if (!r || !g || !b)
    cout << r / 3 + g / 3 + b / 3 << endl;
  else {
    tr = r, tg = g, tb = b;
    mn = min(r, min(g, b));
    r -= mn, g -= mn, b -= mn;
    mn += r / 3 + g / 3 + b / 3;
    ret = mn, mn = 0;
    mn = min(tr, min(tg, tb)) - 1;
    tr -= mn, tg -= mn, tb -= mn;
    mn += tr / 3 + tg / 3 + tb / 3;
    cout << max(mn, ret) << endl;
  }
  return 0;
}

```