#include <bits/stdc++.h>
using namespace std;
void Heavy_Gaurav_Singh() {
  vector<long long> c(3);
  for (int i = 0; i < 3; i++) cin >> c[i];
  long long ans = 0;
  for (int i = 0; i < 3; i++) {
    if (*min_element(c.begin(), c.end()) >= i) {
      long long t = i;
      for (auto x : c) t += (x - i) / 3;
      ans = max(ans, t);
    }
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int test = 1;
  while (test--) {
    Heavy_Gaurav_Singh();
  }
  return 0;
}

### 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;
void Heavy_Gaurav_Singh() {
  vector<long long> c(3);
  for (int i = 0; i < 3; i++) cin >> c[i];
  long long ans = 0;
  for (int i = 0; i < 3; i++) {
    if (*min_element(c.begin(), c.end()) >= i) {
      long long t = i;
      for (auto x : c) t += (x - i) / 3;
      ans = max(ans, t);
    }
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int test = 1;
  while (test--) {
    Heavy_Gaurav_Singh();
  }
  return 0;
}

```