Cquoj 9906 Little Girl and Maximum XOR

Little Girl and Maximum xortime limit:2000msmemory limit:262144kbthis problem would be judged on codeforces. Original ID: 276D
64-bit integer IO format: %i64d Java class name: any

A little girl loves problems on bitwise operations very much. Here ' s one of them.

You are given Integers  L  and  R . Let ' s consider the values Of   For all pairs of Integers  a  and  b Span class= "Apple-converted-space" >  ( l ≤ a ≤ b ≤ R ). Your task is to find the maximum value among all considered ones.

Expression means applying bitwise excluding or operation to integers x and y. The given operation exists in all modern programming languages, for example, in languages C+ + and Java It is represented as "^", in Pascal -as«xor».

Input

The single line contains space-separated integers  l  and   R   (1≤ l ≤ R ≤10^18).

%lld specifier to read or write 64-bit integers inс++. It is preferred to use the CIN, cout streams or the %i64d specifier.

Output

In a single line print a single integer-the maximum value of pairs of integers a, b (l ≤ a ≤ b ≤ R).

Sample Inputinput

1 2

Output

3

Input

8 16

Output

31

Input

1 1

Output

0

1 /*2 April 22, 2016 00:05:103 ^ for XOR or Operation4 after the table can be a regular answer is either 0 or 2 n times-15 6 to get the maximum XOR value, the value must be 2 n times-17 8 that is, in the binary of L and R, the past is traversed from left to right if it encounters (2 ^ i) &l is 1, (2 ^ i) &r for 09 that is, the number of shapes such as 011111111 and 100000000 must exist between L and R. Ten  One can be described in [L, R] exists in 011111111111 and 10000000000 can obtain the maximum XOR value of 2 n times-1 A  - PS: There will be a case where L is 0 R 1 First, because L < R - */ the  -# include <iostream> -# include <cstdio> -# include <cstring> +# include <algorithm> -# include <queue> +# include <vector> A# include <cmath> at# define INF0x3f3f3f3f - using namespacestd; -  - intMainvoid) - { -     inti; in     Long LongL, R, ans; -      while(cin>>l>>R) { to         //1<<i equivalent to 2 of the i-th square +         //1000 0000 0 9th Digit 2 of 8 square 1<<8 -         //so the 64th 2 of the 63-time Square is 1<<63. the          for(i = the; I >=0; i--){ *             if((l& (1ll<<i)) ^ (r& (1ll<<i))) {//1 for int type will be converted to LL $                  Break;Panax Notoginseng             } -         } theAns = (1ll<< (i+1)) -1;//to add one to the original number and minus 1. +cout<<ans<<Endl; A     } the      +     return 0;  -}

Cquoj 9906 Little Girl and Maximum XOR