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;
import java.io.OutputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author Sanchit M. Bhatnagar (sanchitbhatnagar@gmail.com)
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		Scanner in = new Scanner(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskB solver = new TaskB();
		solver.solve(1, in, out);
		out.close();
	}
}

class TaskB {
  public void solve(int testNumber, Scanner in, PrintWriter out) {
    int R = in.nextInt();
    int G = in.nextInt();
    int B = in.nextInt();

    int ans0 = 0;
    ans0 += R / 3;
    ans0 += G / 3;
    ans0 += B / 3;

    int ans1 = 0;
    if (R > 0 && G > 0 && B > 0) {
      ans1++;
    }
    ans1 += Math.max((R - 1) / 3, 0);
    ans1 += Math.max((G - 1) / 3, 0);
    ans1 += Math.max((B - 1) / 3, 0);

    int ans2 = 0;
    if (R > 1 && G > 1 && B > 1) {
      ans2 += 2;
    }
    ans2 += Math.max((R - 2) / 3, 0);
    ans2 += Math.max((G - 2) / 3, 0);
    ans2 += Math.max((B - 2) / 3, 0);

    out.println(Math.max(ans0, Math.max(ans1, ans2)));

  }
}
