#include <bits/stdc++.h>
using namespace std;
template <class T>
T max(T a, T b, T c) {
  return max(a, max(b, c));
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long r, g, b;
  cin >> r >> g >> b;
  long long ans0 = r / 3 + g / 3 + b / 3;
  long long ans1 = (r - 1) / 3 + (g - 1) / 3 + (b - 1) / 3 + 1;
  long long ans2 = (r - 2) / 3 + (g - 2) / 3 + (b - 2) / 3 + 2;
  long long ans = ans0;
  if (min(min(r, g), b) >= 1) ans = max(ans, ans1);
  if (min(min(r, g), b) >= 2) ans = max(ans, ans2);
  cout << ans << '\n';
  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;
template <class T>
T max(T a, T b, T c) {
  return max(a, max(b, c));
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long r, g, b;
  cin >> r >> g >> b;
  long long ans0 = r / 3 + g / 3 + b / 3;
  long long ans1 = (r - 1) / 3 + (g - 1) / 3 + (b - 1) / 3 + 1;
  long long ans2 = (r - 2) / 3 + (g - 2) / 3 + (b - 2) / 3 + 2;
  long long ans = ans0;
  if (min(min(r, g), b) >= 1) ans = max(ans, ans1);
  if (min(min(r, g), b) >= 2) ans = max(ans, ans2);
  cout << ans << '\n';
  return 0;
}

```