codeforces - 276D【贪心+数学】

     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 integersa 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 integersx and y. The given operation exists in all modern programming languages, for example, in languagesC++ and Java it is represented as "^", inPascal — as «xor».

Input

The single line contains space-separated integers l andr (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 thecin, 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).

Examples

Input

1 2

Output

3

Input

8 16

Output

31

Input

1 1

Output

0

题意：给定 l,r。求l <= a <= b <= r。使得 a^b 最大1

题解：贪心，从最高为往最低位考虑。找到第一个l为0，r为1的位置pos，那么a^b最大值就是 pow(2, pos+1)-1;

           注意预处理出所有的 pow(2, i)，或者使用快速幂，不然会精度损失。

代码：

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <bitset>
#include <vector>
#include <queue>
#include <stack>
#include <cmath>
#include <list>
#include <set>
#include <map>
#define rep(i,a,b) for(int i = a;i <= b;++ i)
#define per(i,a,b) for(int i = a;i >= b;-- i)
#define mem(a,b) memset((a),(b),sizeof((a)))
#define FIN freopen("in.txt","r",stdin)
#define FOUT freopen("out.txt","w",stdout)
#define IO ios_base::sync_with_stdio(0),cin.tie(0)
#define mid ((l+r)>>1)
#define ls (id<<1)
#define rs ((id<<1)|1)
#define N 1000+5
#define INF 0x3f3f3f3f
#define INFF 0x3f3f3f3f3f3f3f
typedef long long ll;
const ll mod = 20071027;
const ll eps = 1e-12;
using namespace std;

ll l,r,dp[63];
int main()
{
    dp[1] = 1;
    rep(i, 2, 62)    dp[i] = dp[i-1]*2;
    while(cin >> l >> r){
        int id = -1,bit = 1;
        while(l || r){
            if(r&1 && !(l&1))    id = bit;
            bit++;
            l >>= 1;
            r >>= 1;
        }
        cout << (id == -1 ? 0 : dp[id+1]-1) << endl;
    }
    return 0;
}