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 Main {
    public static void main(String []args) {
        new Main().run();
    }
    
    void run() {
        Scanner sc = new Scanner(System.in);
        int r = sc.nextInt();
        int g = sc.nextInt();
        int b = sc.nextInt();
        
        int ans = Integer.MIN_VALUE;
        // 0, 1, 2, 3 => r, g, b, m
        for(int i = 0; i < 4; i++) {
            for(int j = 0; j < 4; j++) {
                for(int k = 0; k < 4; k++) {
                    for(int l = 0; l < 4; l++) {
                        int temp = 0;
                        int[] a = new int[] {r, g, b};
                        temp += process(a, i);
                        temp += process(a, j);
                        temp += process(a, k);
                        temp += process(a, l);
                        ans = Math.max(ans, temp);
                    }
                }
            }
        }
        
        System.out.println(ans);
    }
    
    int process(int[] a, int operation) {
        int ret = 0;
        if(operation == 3) {
            ret = Math.min(a[0], Math.min(a[1], a[2]));
            a[0] -= ret;
            a[1] -= ret;
            a[2] -= ret;
        } else {
            ret = a[operation] / 3;
            a[operation] %= 3;
        }
        return ret;
    }
}