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.Arrays;
import java.util.Scanner;


public class Main {

	public static void main(String[] args) {
		Scanner in=new Scanner(System.in);
		int[] arr=new int[3];
		for(int i=0;i<arr.length;i++){
			arr[i]=in.nextInt();
		}
		Arrays.sort(arr);
		int cnt=arr[0]+(arr[1]-arr[0])/3+(arr[2]-arr[0])/3;
		int  cnt1 = 0,cnt2=0,cnt3=0,k;
		if(arr[0]>=3){
			k=arr[0]-1;
			if(k>=0)
				cnt1=k+(arr[0]-k)/3+(arr[1]-k)/3+(arr[2]-k)/3;
			if(cnt<cnt1)
				cnt=cnt1;
			k=arr[0]-2;
			if(k>=0)
				cnt2=k+(arr[0]-k)/3+(arr[1]-k)/3+(arr[2]-k)/3;
			if(cnt<cnt2)
				cnt=cnt2;
			k=arr[0]-3;
			if(k>=0)
				cnt3=k+(arr[0]-k)/3+(arr[1]-k)/3+(arr[2]-k)/3;
			if(cnt<cnt3)
				cnt=cnt3;
		}
		System.out.println(cnt);
	}

}

     	 				 	  	 			   		 		 	