#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long int n, x, ans = 0, r, g, b;
  cin >> r >> g >> b;
  if (r == 0 || g == 0 || b == 0) {
    cout << (r / 3) + (g / 3) + (b / 3);
    return 0;
  }
  long long int r1 = r, g1 = g, b1 = b;
  if (r == g && g == b) {
    cout << (r + g + b) / 3;
    return 0;
  }
  long long int mini = min(r, min(g, b));
  while (r > mini) {
    r = r - 3;
    ans++;
  }
  while (g > mini) {
    g = g - 3;
    ans++;
  }
  while (b > mini) {
    b = b - 3;
    ans++;
  }
  ans = ans + min(r, min(g, b));
  ans = max(ans, mini);
  r1 = r1 - mini;
  g1 = g1 - mini;
  b1 = b1 - mini;
  x = mini + (r1 / 3) + (g1 / 3) + (b1 / 3);
  ans = max(ans, x);
  cout << ans;
}

### Prompt

Your challenge is to write 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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long int n, x, ans = 0, r, g, b;
  cin >> r >> g >> b;
  if (r == 0 || g == 0 || b == 0) {
    cout << (r / 3) + (g / 3) + (b / 3);
    return 0;
  }
  long long int r1 = r, g1 = g, b1 = b;
  if (r == g && g == b) {
    cout << (r + g + b) / 3;
    return 0;
  }
  long long int mini = min(r, min(g, b));
  while (r > mini) {
    r = r - 3;
    ans++;
  }
  while (g > mini) {
    g = g - 3;
    ans++;
  }
  while (b > mini) {
    b = b - 3;
    ans++;
  }
  ans = ans + min(r, min(g, b));
  ans = max(ans, mini);
  r1 = r1 - mini;
  g1 = g1 - mini;
  b1 = b1 - mini;
  x = mini + (r1 / 3) + (g1 / 3) + (b1 / 3);
  ans = max(ans, x);
  cout << ans;
}

```