#include <bits/stdc++.h>
using namespace std;
int main() {
  long long r, g, b, ans, r1, g1, b1;
  ans = 0;
  cin >> r >> g >> b;
  r1 = r / (long long)3;
  b1 = b / (long long)3;
  g1 = g / (long long)3;
  ans = r1 + b1 + g1;
  r = r % 3;
  b = b % 3;
  g = g % 3;
  int minn = min(r, min(g, b));
  ans += minn;
  r = r - minn;
  b = b - minn;
  g = g - minn;
  if (r == 0 && r1 > 0) {
    if (min(b, g) == 2) ans++;
  } else if (g == 0 && g1 > 0) {
    if (min(r, b) == 2) ans++;
  } else if (b1 > 0) {
    if (min(r, g) == 2) ans++;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Create a solution in Cpp for 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() {
  long long r, g, b, ans, r1, g1, b1;
  ans = 0;
  cin >> r >> g >> b;
  r1 = r / (long long)3;
  b1 = b / (long long)3;
  g1 = g / (long long)3;
  ans = r1 + b1 + g1;
  r = r % 3;
  b = b % 3;
  g = g % 3;
  int minn = min(r, min(g, b));
  ans += minn;
  r = r - minn;
  b = b - minn;
  g = g - minn;
  if (r == 0 && r1 > 0) {
    if (min(b, g) == 2) ans++;
  } else if (g == 0 && g1 > 0) {
    if (min(r, b) == 2) ans++;
  } else if (b1 > 0) {
    if (min(r, g) == 2) ans++;
  }
  cout << ans << endl;
  return 0;
}

```