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.
import java.util.Scanner;
public class CeilAndFlower {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        long r=sc.nextLong();
        long b=sc.nextLong();
        long g=sc.nextLong();
        
        long m=Math.min(r,Math.min(b, g));
        long c=0;
        //System.out.println(m);
        if(m==0)
        {
            c=r/3 +b/3 +g/3;
            System.out.println(c);
            return;
        }
        if((r%m==2 || b%m==2 || g%m==2) || ((r-m)%3==2 || (b-m)%3==2 || (g-m)%3==2))
        {
            c=m-1;
            r=r-c;
            b=b-c;
            g=g-c;
            
            c=c+(r/3)+(b/3)+(g/3);
            System.out.println(c);
        }
        else
        {
            c=m;
            r=r-c;
            b=b-c;
            g=g-c;
            
            c=c+(r/3)+(b/3)+(g/3);
            System.out.println(c);
        }
        
    }
}
