Good Bye 2015 B. New Year and Old Property 计数问题

B. New Year and Old Property

 

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.

Sample test(s)

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：  算出 a,b 中 二进制下，含有 且 只含有1个0的数 的数目

题解：   对于  10110010  枚举每一位为0所带来的答案就好了

  或者你可以数位dp

//meek
#include<bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include<map>
using namespace std ;
typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
#define pb push_back
#define fi first
#define se second
#define MP make_pair

const int N=400+100;
const ll INF = 1ll<<61;
const int inf = 1000000007;
const int MOD= 1000007;

int d[N];
int sum[N];
ll solve(ll n) {
   if(n<0) return 0;
   ll tmp = n;
   int len = 0;   mem(sum);
   while(tmp)  d[++len]= tmp%2,tmp/=2;
   for(int i=len;i>=1;i--) sum[i] = sum[i+1] +d[i];

   ll ans = 0;
   int f= 0;
   for(int i=1;i<=len;i++) {
    if((!f||d[i])&&sum[i+1]==len-i) {
        ans+=(len-i);//cout<<ans<<endl;
    }
    else ans+=(len-i==0?0:(len-i-1));
     if(d[i] == 0)f=1;
    // cout<<ans<<endl;
   }
   return ans;
}
int main() {
    ll a,b;
     scanf("%lld%lld",&a,&b);
     printf("%lld
",solve(b)-solve(a-1));
   //  printf("%lld
",);
    return 0;
}