Codeforces 768B Code For 1

Jon fought bravely to rescue the wildlings who were attacked by the white-walkers at Hardhome. On his arrival, Sam tells him that he wants to go to Oldtown to train at the Citadel to become a maester, so he can return and take the deceased Aemon's place as maester of Castle Black. Jon agrees to Sam's proposal and Sam sets off his journey to the Citadel. However becoming a trainee at the Citadel is not a cakewalk and hence the maesters at the Citadel gave Sam a problem to test his eligibility.

Initially Sam has a list with a single element n. Then he has to perform certain operations on this list. In each operation Sam must remove any element x, such that x > 1, from the list and insert at the same position $\text{[math]}$ , $\text{[math]}$ , $\text{[math]}$ sequentially. He must continue with these operations until all the elements in the list are either 0 or 1.

Now the masters want the total number of 1s in the range l to r (1-indexed). Sam wants to become a maester but unfortunately he cannot solve this problem. Can you help Sam to pass the eligibility test?

Input

The first line contains three integers n, l, r (0 ≤ n < 2⁵⁰, 0 ≤ r - l ≤ 10⁵, r ≥ 1, l ≥ 1) – initial element and the range l to r.

It is guaranteed that r is not greater than the length of the final list.

Output

Output the total number of 1s in the range l to r in the final sequence.

Examples

Input

7 2 5

Output

Input

10 3 10

Output

Note

Consider first example:

$\text{[math]}$

Elements on positions from 2-nd to 5-th in list is [1, 1, 1, 1]. The number of ones is 4.

For the second example:

$\text{[math]}$

Elements on positions from 3-rd to 10-th in list is [1, 1, 1, 0, 1, 0, 1, 0]. The number of ones is 5.

题目链接：http://codeforces.com/contest/768/problem/B

分析：二分的模版题！来围观看看！

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 ll n, l, r, s = 1, ans;
 5 void solve(ll a, ll b, ll l, ll r, ll d)//二分的思想
 6 {
 7     if ( a > b || l > r )    return;
 8     else
 9         {
10         ll mid = (a+b)/2;
11         if ( r < mid )solve(a,mid-1,l,r,d/2);
12         else if ( mid < l )solve(mid+1,b,l,r,d/2);
13         else {
14             ans += d%2;
15         solve(a,mid-1,l,mid-1,d/2);
16         solve(mid+1,b,mid+1,r,d/2);
17         }
18     }
19 }
20 int main()
21 {
22     cin >> n >> l >> r;
23     long long p = n;
24     while ( p >= 2 )
25     {
26         p /= 2;
27         s = s*2+1;
28     }
29     solve(1,s,l,r,n);
30     cout << ans << endl;
31     return 0;
32 }