#include <bits/stdc++.h>
using namespace std;
int main() {
  long long a, b, c;
  cin >> a >> b >> c;
  vector<long long> v, z;
  v.push_back(a);
  v.push_back(b);
  v.push_back(c);
  sort(v.begin(), v.end());
  long long cnt = a / 3;
  a = a % 3;
  cnt = cnt + b / 3;
  b = b % 3;
  cnt = cnt + c / 3;
  c = c % 3;
  long long x = min(a, (min(b, c)));
  cnt = cnt + x;
  a = a - x;
  b = b - x;
  c = c - x;
  z.push_back(a);
  z.push_back(b);
  z.push_back(c);
  sort(z.begin(), z.end());
  if (z[0] == 0 && z[1] == 2 && z[2] == 2 && v[0] != 0) cnt++;
  cout << cnt;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
  cin >> a >> b >> c;
  vector<long long> v, z;
  v.push_back(a);
  v.push_back(b);
  v.push_back(c);
  sort(v.begin(), v.end());
  long long cnt = a / 3;
  a = a % 3;
  cnt = cnt + b / 3;
  b = b % 3;
  cnt = cnt + c / 3;
  c = c % 3;
  long long x = min(a, (min(b, c)));
  cnt = cnt + x;
  a = a - x;
  b = b - x;
  c = c - x;
  z.push_back(a);
  z.push_back(b);
  z.push_back(c);
  sort(z.begin(), z.end());
  if (z[0] == 0 && z[1] == 2 && z[2] == 2 && v[0] != 0) cnt++;
  cout << cnt;
}

```