New Year and Old Property CodeForces - 611B

The year 2015 is almost over.

Limak is a little polar bear. He has recently learnt about the binary system. He noticed that the passing year has exactly one zero in its representation in the binary system — 201510 = 111110111112. Note that he doesn't care about the number of zeros in the decimal representation.

Limak chose some interval of years. He is going to count all years from this interval that have exactly one zero in the binary representation. Can you do it faster?

Assume that all positive integers are always written without leading zeros.

The only line of the input contains two integers a and b (1 ≤ a ≤ b ≤ 1018) — the first year and the last year in Limak's interval respectively.

Print one integer – the number of years Limak will count in his chosen interval.

5 10

Output

2

Input

2015 2015

Output

1

Input

100 105

Output

0

Input

72057594000000000 72057595000000000

Output

26

In the first sample Limak's interval contains numbers 510 = 1012, 610 = 1102, 710 = 1112, 810 = 10002, 910 = 10012 and 1010 = 10102. Two of them (1012 and 1102) have the described property.

#include<bits/stdc++.h>
using namespace std;
typedef long long  ll;
ll top[100];
inline void  find_top()
{
int i;
top[0]=1;
for (i=1;i<=64;i++)
{
top[i]=top[i-1]+pow(2,i);
}
}
int solve (ll &a,ll &b)
{
int i,j;
int ans =0;
for (i=0;i<64;i++)
{
ll temp=pow(2,i);
for (j=i+1;j<64;j++)
if (top[j]-temp>=a&&top[j]-temp<=b)
ans++;
}
return ans;
}

int main ()
{
find_top();
ll a, b;
cin>>a>>b;
cout<<solve(a,b)<<endl;

return 0;
}