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.*;

public class D{
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int r = sc.nextInt();
		int g = sc.nextInt();
		int b = sc.nextInt();
		int tempR = r;
		int tempG = g;
		int tempB = b;
		int ans = 0;
		while(tempR % 3 != 0 && tempR>0 && tempG>0 && tempB>0) {
			tempR--;
			tempG--;
			tempB--;
			ans++;
		}
		ans += tempR / 3;
		ans += tempG / 3;
		ans += tempB / 3;
		
		tempR = r;
		tempG = g;
		tempB = b;
		int tempAns = 0;
		while(tempG % 3 != 0 && tempR>0 && tempG>0 && tempB>0) {
			tempR--;
			tempG--;
			tempB--;
			tempAns++;
		}
		tempAns += tempR / 3;
		tempAns += tempG / 3;
		tempAns += tempB / 3;
		ans = Math.max(ans, tempAns);
		
		tempR = r;
		tempG = g;
		tempB = b;
		tempAns = 0;
		while(tempB % 3 != 0 && tempR>0 && tempG>0 && tempB>0) {
			tempR--;
			tempG--;
			tempB--;
			tempAns++;
		}
		tempAns += tempR / 3;
		tempAns += tempG / 3;
		tempAns += tempB / 3;
		ans = Math.max(ans, tempAns);
		System.out.println(ans);
	}
}