Codeforces 276D Little Girl and Maximum XOR【贪心】

D. Little Girl and Maximum XOR

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 %I64dspecifier.

Output

In a single line print a single integer — the maximum value of  for all pairs of integers a, b (l ≤ a ≤ b ≤ r).

Examples

input

1 2

output

3

input

8 16

output

31

input

1 1

output

0

题目大意：

在区间【L,R】内找到两个数，使得他们的异或值最大。

思路：

①我们从高位开始贪心即可，如果两个数L,R的2进制位数长度不同，设定R的2进制长度为x的话，那么答案一定是(1<<x)-1；因为我们可以让R除了最高位之外，都变成0，让L所有位子都变成1.并且长度为x-1.

②否则，如果两个数长度相等且部分前边高位数值一样的话，我们忽略不计，只找最高位不同的位子即可。

Ac代码：

#include<stdio.h>#include<string.h>using namespace std;#define ll __int64int main(){    ll l,r;    while(~scanf("%I64d%I64d",&l,&r))    {        for(ll i=60;i>=-1;i--)        {            if(i==-1||((l>>i)^(r>>i)))            {                printf("%I64d\n",((1ll<<(i+1))-1));                break;            }        }    }}