Problem:
Score : 900 points

Problem Statement

Nukes has an integer that can be represented as the bitwise OR of one or more integers between A and B (inclusive).
How many possible candidates of the value of Nukes's integer there are?

Constraints

1 ≤ A ≤ B < 2^{60}

A and B are integers.

Input

The input is given from Standard Input in the following format:

A
B

Output

Print the number of possible candidates of the value of Nukes's integer.

Sample Input 1

7
9

Sample Output 1

4

In this case, A=7 and B=9. There are four integers that can be represented as the bitwise OR of a non-empty subset of {7, 8, 9}: 7, 8, 9 and 15.

Sample Input 2

65
98

Sample Output 2

63

Sample Input 3

271828182845904523
314159265358979323

Sample Output 3

68833183630578410

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author majk
 */

#ifndef MAJK_LIB
#define MAJK_LIB

#include <vector>
#include <stack>
#include <iostream>
#include <unordered_map>
#include <unordered_set>
#include <map>
#include <iomanip>
#include <set>
#include <functional>
#include <algorithm>
#include <numeric>
#include <cassert>
#include <cmath>
#include <string>
#include <queue>
#include <array>
#include <bitset>
using namespace std;

#define x first
#define y second
typedef std::pair<int,int> pii; typedef long long ll; typedef unsigned long long ull; typedef unsigned int ui; typedef pair<ui,ui> puu;

template <typename T, typename U> std::istream&operator>>(std::istream&i, pair<T,U>&p) {i >> p.x >> p.y; return i;}
template<typename T>std::istream&operator>>(std::istream&i,vector<T>&t) {for(auto&v:t){i>>v;}return i;}
template <typename T, typename U> std::ostream&operator<<(std::ostream&o, const pair<T,U>&p) {o << p.x << ' ' << p.y; return o;}
template<typename T>std::ostream&operator<<(std::ostream&o,const vector<T>&t) {if(t.empty())o<<'\n';for(size_t i=0;i<t.size();++i){o<<t[i]<<" \n"[i == t.size()-1];}return o;}
template<typename T> using minheap = priority_queue<T, vector<T>, greater<T>>;
template<typename T> using maxheap = priority_queue<T, vector<T>, less<T>>;
ui logceil(ll x) { return x?8*sizeof(ll)-__builtin_clzll(x):0; }

namespace std { template<typename T,typename U>struct hash<pair<T,U>>{hash<T>t;hash<U>u;size_t operator()(const pair<T,U>&p)const{return t(p.x)^(u(p.y)<<7);}}; }
template<typename T,typename F>T bsh(T l,T h,const F&f){T r=-1,m;while(l<=h){m=(l+h)/2;if(f(m)){l=m+1;r=m;}else{h=m-1;}}return r;}
template<typename F> double bshd(double l,double h,const F&f,double p=1e-9){ui r=3+(ui)log2((h-l)/p);while(r--){double m=(l+h)/2;if(f(m)){l=m;}else{h=m;}}return (l+h)/2;}
template<typename T,typename F>T bsl(T l,T h,const F&f){T r=-1,m;while(l<=h){m=(l+h)/2;if(f(m)){h=m-1;r=m;}else{l=m+1;}}return r;}
template<typename F> double bsld(double l,double h,const F&f,double p=1e-9){ui r=3+(ui)log2((h-l)/p);while(r--){double m=(l+h)/2;if(f(m)){h=m;}else{l=m;}}return (l+h)/2;}
template<typename T> T gcd(T a,T b) { if (a<b) swap(a,b); return b?gcd(b,a%b):a; }


template<typename T>class vector2:public vector<vector<T>>{public:vector2(){} vector2(size_t a,size_t b,T t=T()):vector<vector<T>>(a,vector<T>(b,t)){}};
template<typename T>class vector3:public vector<vector2<T>>{public:vector3(){} vector3(size_t a,size_t b,size_t c,T t=T()):vector<vector2<T>>(a,vector2<T>(b,c,t)){}};
template<typename T>class vector4:public vector<vector3<T>>{public:vector4(){} vector4(size_t a,size_t b,size_t c,size_t d,T t=T()):vector<vector3<T>>(a,vector3<T>(b,c,d,t)){}};
template<typename T>class vector5:public vector<vector4<T>>{public:vector5(){} vector5(size_t a,size_t b,size_t c,size_t d,size_t e,T t=T()):vector<vector4<T>>(a,vector4<T>(b,c,d,e,t)){}};


#endif

class DAOrorBProblem {
public:
    void solve(istream& cin, ostream& cout) {
        ll A, B;
        cin >> A >> B;
        if (A == B) {
            cout << 1 << endl;
            return;
        }

        for (ll l = 1LL<<60; l >= 1; l >>= 1) {
            if ((A&l) == (B&l)) {
                A &= ~(l);
                B &= ~(l);
            } else {
                B &= ~(l);
                ll k = l;
                while (k >= 1 && (B&k) == 0) k >>= 1;
                ll lo = A;
                ll hi = k == 0 ? 0 : (2*k)-1;
                cout << 2 * l - lo + min(0LL, - lo + hi + 1) << endl;
                break;
            }
        }
    }
};


int main() {
	ios_base::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
	DAOrorBProblem solver;
	std::istream& in(std::cin);
	std::ostream& out(std::cout);
	solver.solve(in, out);
    return 0;
}