#include <bits/stdc++.h>
using namespace std;
long long a[3];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long ans = 0;
  for (int i = 0; i < 3; i++) {
    cin >> a[i];
    long long p = max(0LL, (a[i] + 2) / 3 - 1);
    ans += p;
    a[i] -= p * 3;
  }
  sort(a, a + 3);
  if (a[0] == 0)
    cout << ans + a[1] / 3 + a[2] / 3 << '\n';
  else if (a[0] == 1 && a[1] == 2 && a[2] == 3)
    cout << ans + 1 << '\n';
  else
    cout << ans + (a[0] + a[1] + a[2]) / 3 << '\n';
}

### 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;
long long a[3];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long ans = 0;
  for (int i = 0; i < 3; i++) {
    cin >> a[i];
    long long p = max(0LL, (a[i] + 2) / 3 - 1);
    ans += p;
    a[i] -= p * 3;
  }
  sort(a, a + 3);
  if (a[0] == 0)
    cout << ans + a[1] / 3 + a[2] / 3 << '\n';
  else if (a[0] == 1 && a[1] == 2 && a[2] == 3)
    cout << ans + 1 << '\n';
  else
    cout << ans + (a[0] + a[1] + a[2]) / 3 << '\n';
}

```