#include <bits/stdc++.h>
using namespace std;
long long int hell = 1000000000;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int a[3];
  cin >> a[0] >> a[1] >> a[2];
  sort(a, a + 3);
  long long int p = 0, q = 0;
  q = hell;
  p = a[0] / 3 + a[2] / 3 + a[1] / 3;
  q = min(q, (a[0] % 3));
  q = min(q, (a[1] % 3));
  q = min(q, (a[2] % 3));
  p = p + q;
  if (a[0] > 0 && p > 0 && a[0] % 3 == 0 && a[1] % 3 == 2 && a[2] % 3 == 2) p++;
  if (p > 0 && a[1] % 3 == 0 && a[0] % 3 == 2 && a[2] % 3 == 2) p++;
  if (p > 0 && a[2] % 3 == 0 && a[1] % 3 == 2 && a[0] % 3 == 2) p++;
  cout << p;
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 hell = 1000000000;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int a[3];
  cin >> a[0] >> a[1] >> a[2];
  sort(a, a + 3);
  long long int p = 0, q = 0;
  q = hell;
  p = a[0] / 3 + a[2] / 3 + a[1] / 3;
  q = min(q, (a[0] % 3));
  q = min(q, (a[1] % 3));
  q = min(q, (a[2] % 3));
  p = p + q;
  if (a[0] > 0 && p > 0 && a[0] % 3 == 0 && a[1] % 3 == 2 && a[2] % 3 == 2) p++;
  if (p > 0 && a[1] % 3 == 0 && a[0] % 3 == 2 && a[2] % 3 == 2) p++;
  if (p > 0 && a[2] % 3 == 0 && a[1] % 3 == 2 && a[0] % 3 == 2) p++;
  cout << p;
  return 0;
}

```