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.
//package codeforces;

import java.util.Scanner;

/**
 * Created by nitin.s on 15/01/17.
 */
public class CielAndFlowers {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int r = in.nextInt(), g = in.nextInt(), b = in.nextInt();
        int answer = 0;
        for(int i = 0; i < 3; ++i) {
            if(i == 0) {
                int tmp = 0;
                tmp += r / 3;
                tmp += g / 3;
                tmp += b / 3;
                answer = Math.max(answer, tmp);
            } else if(i == 1) {
                int tmp = 0;
                int minn = Math.min(r, Math.min(g, b));
                tmp += (r - 1) / 3;
                tmp += (g - 1) / 3;
                tmp += (b - 1) / 3;
                if(minn >= 1)
                    tmp += 1;
                answer = Math.max(answer, tmp);
            } else if(i == 2) {
                int tmp = 0;
                int minn = Math.min(r, Math.min(g, b));
                tmp += (r - 2) / 3;
                tmp += (g - 2) / 3;
                tmp += (b - 2) / 3;
                if(minn >= 2)
                    tmp += 2;
                answer = Math.max(answer, tmp);
            }
        }
        System.out.println(answer);
    }
}
