l=[]
for i in range(2,100):
k=1
while(2**i-1-k>0 and k<2**(i-1)):
l.append((2**i)-1-k)
k*=2
n,m=map(int, input().split())
nb=0
for j in l:
if (j>=n) and (j<=m) :
nb+=1
print(nb)#include<bits/stdc++.h>
#define int unsigned long long
#define fa(i,a,n) for(int i=a;i<n;i++)
#define pb push_back
#define bp pop_back
#define mp make_pair
#define all(v) v.begin(),v.end()
#define vi vector<int>
#define Utaval ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
using namespace std;
//int arr[200010][18];
vector<int>v;
void bina(int n,int ind){
for(int i=0;i<=ind;i++){
int numbery= n-(int)pow(2,i);
v.pb(numbery);
}
}
void solve(){
int a,b,ans=0;
cin>>a>>b;
fa(i,0,v.size()){
if(v[i]>=a && v[i]<=b)ans++;
}
cout<<ans<<endl;
}
signed main()
{
Utaval;
long long t=1;
//cin>>t;
for(int i=1;i<=64;i++){
int ans = pow(2,i+1)-1;
bina(ans,i-1);
}
sort(v.begin(),v.end());
while(t--)
solve();
}
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