Description

Alice is exploring the wonderland, suddenly she fell into a hole, when she woke up, she found there are b - a + 1 treasures labled a from bin front of her.

Alice was very excited but unfortunately not all of the treasures are real, some are fake.

Now we know a treasure labled n is real if and only if [n/1] + [n/2] + ... + [n/k] + ... is even.

Now given 2 integers a and b, your job is to calculate how many real treasures are there.

Input

The input contains multiple cases, each case contains two integers a and b (0 <= a <= b <= 2⁶³-1) seperated by a single space. Proceed to the end of file.

Output

Output the total number of real treasure.

Sample Input

0 20 10

Sample Output

16 *********************************************************************************************************************************************************** 通过计算我们会发现，只有在区间[0,1)、[4,9)、[16,25)……实属满足题意 **********************************************************************************************************************************************************

1 #include
        
          2 #include
         
           3 #include
          
            4 #include
           
             5 #include
            
              6 #include
             
               7 using namespace std; 8 typedef long long ll; 9 const double eps=1e-6;10 ll solve(ll n)11 {12 if (n == -1) return 0;13 ll r = (ll)sqrt((double) n + 1 + eps);14 if (r & 1){15 r = (r + 1) >> 1;16 return ((r << 1) - 1) * r;17 }18 else{19 return n - r * r + (r >> 1) * (r - 1) + 1;//减去两区间之间的无用数20 }21 }22 int main()23 {24 ll a,b;25 while(~scanf("%lld %lld",&a,&b))26 {27 ll ans=solve(b)-solve(a-1);28 printf("%lld\n",ans);29 }30 return 0;31 }