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() {
  long long int n, a, b, c, i, j, k, x, y, z, x1, y1, z1;
  cin >> a >> b >> c;
  x = a % 3;
  y = b % 3;
  z = c % 3;
  x1 = a / 3;
  y1 = b / 3;
  z1 = c / 3;
  long long int total = x1 + y1 + z1;
  if (x == 1 && y == 1 && z == 1)
    cout << total + 1;
  else if (x == 2 && y == 2 && z == 2)
    cout << total + 2;
  else if ((x == y && x == 1 && z == 2) || (x == z && x == 1 && y == 2) ||
           (z == y && z == 1 && x == 2))
    cout << total + 1;
  else if ((x == y && x == 2 && z == 1) || (x == z && x == 2 && y == 1) ||
           (z == y && z == 2 && x == 1))
    cout << total + 1;
  else if ((x == y && x == 2 && z == 0 && c > 0) ||
           (x == z && x == 2 && y == 0 && b > 0) ||
           (z == y && z == 2 && x == 0 && a > 0))
    cout << total + 1;
  else
    cout << total;
  return 0;
}
