从左边界往右找第一个%7=1的数x
从右边界往左找第一个%7=1的数y
然后用(y/x)+1即为里面符合条件的整数的个数

import java.math.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.*;
public class Main {
    public static void main(String args[])throws IOException
    {
        StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        in.nextToken();
        int a = (int)in.nval;
       in.nextToken();
        int b = (int)in.nval;
        if((long)(checkzuo((long)Math.pow(2,b)-1)-checkyou((long)Math.pow(2,a)))/7<0)
            out.println(0);
        else
            out.print(((long)(checkzuo((long)Math.pow(2,b)-1)-checkyou((long)Math.pow(2,a)))/7)+1);
        out.flush();
    }
    public static long checkzuo(long a)
    {
        long x = a;
        while(x%7!=1)
        {
            x--;
        }
        return x;
    }
    public static long checkyou(long a)
    {
        long x = a;
        while(x%7!=1)
        {
            x++;
        }
        return x;
    }
                  }