#include <bits/stdc++.h>
using namespace std;
int main() {
  long long a, b, c, sum, s1, s2, s3, mn;
  cin >> a >> b >> c;
  mn = min(a, min(c, b));
  sum = mn + (a - mn) / 3 + (b - mn) / 3 + (c - mn) / 3;
  long long p = mn - 1;
  if (p >= 0) {
    s1 = p + (a - p) / 3 + (b - p) / 3 + (c - p) / 3;
    sum = max(sum, s1);
  }
  p = mn - 2;
  if (p >= 0) {
    s2 = p + (a - p) / 3 + (b - p) / 3 + (c - p) / 3;
    sum = max(s2, sum);
  }
  p = mn - 3;
  if (p >= 0) {
    s2 = p + (a - p) / 3 + (b - p) / 3 + (c - p) / 3;
    sum = max(s2, sum);
  }
  cout << sum << endl;
  return 0;
}

### Prompt

Your challenge is to write 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;
int main() {
  long long a, b, c, sum, s1, s2, s3, mn;
  cin >> a >> b >> c;
  mn = min(a, min(c, b));
  sum = mn + (a - mn) / 3 + (b - mn) / 3 + (c - mn) / 3;
  long long p = mn - 1;
  if (p >= 0) {
    s1 = p + (a - p) / 3 + (b - p) / 3 + (c - p) / 3;
    sum = max(sum, s1);
  }
  p = mn - 2;
  if (p >= 0) {
    s2 = p + (a - p) / 3 + (b - p) / 3 + (c - p) / 3;
    sum = max(s2, sum);
  }
  p = mn - 3;
  if (p >= 0) {
    s2 = p + (a - p) / 3 + (b - p) / 3 + (c - p) / 3;
    sum = max(s2, sum);
  }
  cout << sum << endl;
  return 0;
}

```