#include <bits/stdc++.h>
using namespace std;
long long mod(long long a, long long b) {
  long long ans;
  if (b == 0)
    return -1;
  else {
    ans = (a < 0 ? mod(((a % b) + b), b) : a % b);
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  int c[3];
  for (int i = 0; i <= 2; i++) cin >> c[i];
  int ans = 0;
  for (int i = 0; i <= 2; i++) {
    if (c[0] < i || c[1] < i || c[2] < i) break;
    ans = max(ans, i + (c[0] - i) / 3 + (c[1] - i) / 3 + (c[2] - i) / 3);
  }
  cout << ans << endl;
}

### 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;
long long mod(long long a, long long b) {
  long long ans;
  if (b == 0)
    return -1;
  else {
    ans = (a < 0 ? mod(((a % b) + b), b) : a % b);
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  int c[3];
  for (int i = 0; i <= 2; i++) cin >> c[i];
  int ans = 0;
  for (int i = 0; i <= 2; i++) {
    if (c[0] < i || c[1] < i || c[2] < i) break;
    ans = max(ans, i + (c[0] - i) / 3 + (c[1] - i) / 3 + (c[2] - i) / 3);
  }
  cout << ans << endl;
}

```