codeforces 276D Little Girl and Maximum XOR(区间最大异或值--技巧)【模板】

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

You are given two integers l and r. Let's consider the values of  for all pairs of integers a and b (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 ≤ 10¹⁸).

Please, do not use the %lld specifier to read or write 64-bit integers in С++. 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  for all pairs of integers a, b (l ≤ a ≤ b ≤ r).

Example

Input

1 2

Output

3

Input

8 16

Output

31

Input

1 1

Output

0

【题解】

 刚开始一看到数据范围惊呆了，但是最后仔细思考以后发现是有技巧可言的，首先，异或值要为1，那么两个数的对应2位进制位要不同，而异或值要最大，则二进制的高位要尽可能的为1，所以这就是切入点，从给定的区间上下限入手（l和R），从l和r二进制的最高位开始比较，如果出现对应位异或值位1，就从该处开始，低位都置为1，此时其表示的数就是最大异或值了。

【AC代码】

#include<cstdio>#include<algorithm>#include<cstdio>#include<math.h>#include<string.h>#include<iostream>using namespace std;typedef __int64 ll;ll l,r;int main(){    while(~scanf("%I64d%I64d",&l,&r))    {        ll max_xor = 0;        int i = 61;        for(;i>=0;--i)        {            if(l&(1LL<<i) ^ r&(1LL<<i)) break;        }        for(;i>=0;--i)        {            max_xor|= 1LL<<i;        }        printf("%I64d\n",max_xor);    }    return 0;}