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 Ciel_and_Flowers {
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int r = scan.nextInt();
		int g = scan.nextInt();
		int b = scan.nextInt();
		int min = Math.min(r, Math.min(g, b));
		long max1 = min + ( ((r-min)/3) + ((g-min)/3) + ((b-min)/3) );
		long max2 = (r/3) + (g/3) + (b/3);
		if (r%3 != 0 && g%3 != 0 && b%3 != 0) {
			max2 += Math.min(r%3, Math.min(g%3, b%3));
		}
		
		long max3 = 0;
		if (min == r) {
			if (r <= 3) {
				max3 += g/3 + b/3;
				max3 += Math.min(r, Math.min(g%3, b%3));
			} 
		} else if (min == g) {
			if (g <= 3) {
				max3 += r/3 + b/3;
				max3 += Math.min(r%3, Math.min(g, b%3));
			}
		} else if (min == b) {
			if (b <= 3) {
				max3 += r/3 + g/3;
				max3 += Math.min(r%3, Math.min(g%3, b));
			}
		}
		
		long max4  = 0;
		if (min != 0)
			max4 = min-1 + ( ((r-(min-1))/3) + ((g-(min-1))/3) + ((b-(min-1))/3) );;
		
		
//		System.out.println(max1 + "  " + max2 + "  " + max3);
		System.out.println(Math.max(max1, Math.max(max2, Math.max(max3, max4))));
	}

}
