#include <bits/stdc++.h>
using namespace std;
long long one(long long a, long long b, long long c) {
  long long ans = a / 3 + b / 3 + c / 3;
  a %= 3;
  b %= 3;
  c %= 3;
  ans += min(a, min(b, c));
  return ans;
}
long long two(long long a, long long b, long long c) {
  long long ans = min(a, min(b, c));
  a -= ans;
  b -= ans;
  c -= ans;
  ans += a / 3 + b / 3 + c / 3;
  return ans;
}
int main() {
  long long a, b, c;
  cin >> a >> b >> c;
  long long ans = max(0LL, min(a, min(b, c)) - 1);
  a -= ans;
  b -= ans;
  c -= ans;
  ans += max(one(a, b, c), two(a, b, c));
  cout << ans;
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 one(long long a, long long b, long long c) {
  long long ans = a / 3 + b / 3 + c / 3;
  a %= 3;
  b %= 3;
  c %= 3;
  ans += min(a, min(b, c));
  return ans;
}
long long two(long long a, long long b, long long c) {
  long long ans = min(a, min(b, c));
  a -= ans;
  b -= ans;
  c -= ans;
  ans += a / 3 + b / 3 + c / 3;
  return ans;
}
int main() {
  long long a, b, c;
  cin >> a >> b >> c;
  long long ans = max(0LL, min(a, min(b, c)) - 1);
  a -= ans;
  b -= ans;
  c -= ans;
  ans += max(one(a, b, c), two(a, b, c));
  cout << ans;
  return 0;
}

```