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 CielAndFlowers {
	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		long r = s.nextLong();
		long b = s.nextLong();
		long g = s.nextLong();
		if (r < 3 && b < 3 && g < 3)
			System.out.println(0);
		
		else if(r==0 )
    	   System.out.println((g/3)+(b/3));
		else if(g==0)
			System.out.println((r/3)+(b/3));
		else if(b==0)
			System.out.println((r/3)+(g/3));
		else {

			long remaining = (r / 3) + (b / 3) + (g / 3);
			long remain = Math.max(Math.max(r % 3, b % 3), g % 3);
			remain = ((r - remain) / 3) + ((b - remain) / 3) + ((g - remain) / 3) + remain;
			if (r % 3 != 0 && g % 3 != 0 && b % 3 != 0)
				remaining += ((r % 3) + (b % 3) + (g % 3)) / 3;
			// System.out.println();
			System.out.println(Math.max(remain, Math.max(remaining, Math.min(Math.min(r, g), b))));
		}
	}

}