codeforces611B. New Year and Old Property

B. New Year and Old Property

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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.

Input

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.

Output

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

Examples

input

5 10

output

2

input

2015 2015

output

1

input

100 105

output

0

input

72057594000000000 72057595000000000

output

26

Note

In the first sample Limak's interval contains numbers 510 = 1012, 610 = 1102, 710 = 1112, 810 = 10002, 910 = 10012 and1010 = 10102.
Two of them (1012 and 1102)
have the described property.

题意：给出两个数求在这个区间中的数用二进制表示下只有1个0的数个个数

解题思路：计算出两个的二进制位数然后对位数在其之间的直接计算边缘的dfs即可

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<list>
#include<vector>
using namespace std;
int Count(long long n){
int cnt=0;
while(n){
n>>=1;
cnt++;
}
return cnt;
}
long long a,b,ans;
void dfs(long long num,bool sign,int cnt,int std){
if(cnt==std){
if(num>=a&&num<=b&&sign)ans++;
return ;
}
if(!sign)dfs(num*2,true,cnt+1,std);
dfs(num*2+1,sign,cnt+1,std);
}
int main()
{
scanf("%lld%lld",&a,&b);
int cnta=Count(a),cntb=Count(b);
if(cnta==cntb){
ans=0;
dfs(1,false,1,cnta);
printf("%lld\n",ans);
}
else {
ans=0;
dfs(1,false,1,cnta);
dfs(1,false,1,cntb);
for(int i=cnta+1;i<=cntb-1;++i){
ans+=i-1;
}
printf("%lld\n",ans);
}
return 0;
}