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.
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long int r, b, g, c = 0, v, z;
  cin >> r >> g >> b;
  if ((r + b + g) < 3) {
    cout << "0";
  } else {
    if (r == 0 || b == 0 || g == 0) {
      c = (r / 3) + (b / 3) + (g / 3);
      cout << c;
    } else if (((r % 3) == 0 || (b % 3) == 0) &&
               ((r % 3) == 0 || (g % 3) == 0) &&
               ((b % 3) == 0 || (g % 3) == 0)) {
      c = (r / 3) + (b / 3) + (g / 3);
      cout << c;
    } else {
      if (r % 3 == 0)
        z = r;
      else if (b % 3 == 0)
        z = b;
      else
        z = g;
      if (r < b && r < g)
        if (r != z)
          v = r % 3;
        else if (b > g)
          v = b % 3;
        else
          v = g % 3;
      else if (b < r && b < g)
        if (b % 3 != 0)
          v = b % 3;
        else if (r < g)
          v = r % 3;
        else
          v = g % 3;
      else if (g % 3 != 0)
        v = g % 3;
      else if (b < r)
        v = b % 3;
      else
        v = r % 3;
      r -= v;
      b -= v;
      g -= v;
      c = (r / 3) + (b / 3) + (g / 3);
      cout << c + v;
    }
  }
  return 0;
}
