#include <bits/stdc++.h>
using namespace std;
using namespace std;
int main() {
  long long r, g, b;
  scanf("%lld", &r);
  scanf("%lld", &g);
  scanf("%lld", &b);
  long long r1 = r, g1 = g, b1 = b;
  long long ans = 0;
  ans += min(min(r, g), b);
  r -= ans;
  g -= ans;
  b -= ans;
  long long ans1 = ans - 1;
  ans += r / 3;
  ans += g / 3;
  ans += b / 3;
  r++;
  g++;
  b++;
  if (ans1 > 0) {
    ans1 += r / 3;
    ans1 += g / 3;
    ans1 += b / 3;
    if (ans1 > ans) ans = ans1;
  }
  cout << ans << endl;
}

### Prompt

Generate 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;
using namespace std;
int main() {
  long long r, g, b;
  scanf("%lld", &r);
  scanf("%lld", &g);
  scanf("%lld", &b);
  long long r1 = r, g1 = g, b1 = b;
  long long ans = 0;
  ans += min(min(r, g), b);
  r -= ans;
  g -= ans;
  b -= ans;
  long long ans1 = ans - 1;
  ans += r / 3;
  ans += g / 3;
  ans += b / 3;
  r++;
  g++;
  b++;
  if (ans1 > 0) {
    ans1 += r / 3;
    ans1 += g / 3;
    ans1 += b / 3;
    if (ans1 > ans) ans = ans1;
  }
  cout << ans << endl;
}

```