【Codeforces Round 169 (Div 2) D】【簡單數位貪心】Little Girl and Maximum XOR 區間選兩數使得異或值儘可能大

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

#include<stdio.h> 
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 0, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int n;
LL L, R;
int main()
{
	while (~scanf("%lld%lld", &L, &R))
	{
		LL ans = 0;
		for (int k = 60; k >= 0; --k)
		{
			int a = L >> k & 1;
			int b = R >> k & 1;
			if (a == b);
			else
			{
				ans = (1ll << (k + 1)) - 1;
				break;
			}
		}
		printf("%lld\n", ans);
	}
	return 0;
}
/*
【題意】
告訴你一個大區間[L, R]
然你找出其中的數字x、y，滿足[L <= x,y <= R]並使得x ^ y的值儘可能大

【分析】
我們假設大的數與R在前面若干位保持一致，小的數與L在前面若干位保持一致。
那這兩個數的最高分歧位是第k位，這一位上，大數是1，小數是0，則在此之後，大數可以一直取0，小數可以一直取1。
於是，答案就是(1ll << (k + 1)) - 1;

*/