Problem:
Score : 200 points

Problem Statement

There are N apple trees in a row. People say that one of them will bear golden apples.

We want to deploy some number of inspectors so that each of these trees will be inspected.

Each inspector will be deployed under one of the trees. For convenience, we will assign numbers from 1 through N to the trees. An inspector deployed under the i-th tree (1 \leq i \leq N) will inspect the trees with numbers between i-D and i+D (inclusive).

Find the minimum number of inspectors that we need to deploy to achieve the objective.

Constraints

All values in input are integers.

1 \leq N \leq 20

1 \leq D \leq 20

Input

Input is given from Standard Input in the following format:

N D

Output

Print the minimum number of inspectors that we need to deploy to achieve the objective.

Sample Input 1

6 2

Sample Output 1

2

We can achieve the objective by, for example, placing an inspector under Tree 3 and Tree 4.

Sample Input 2

14 3

Sample Output 2

2

Sample Input 3

20 4

Sample Output 3

3

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

//--------ROMAN ITERATION -------------
#include <bits/stdc++.h>
#define ll  long long int
#define pb push_back
#define ff first
#define ss second
#define vi vector<int>
#define br cout <<"\n";
#define all(x)  (x).begin(),(x).end()
#define rall(x)  (x).rbegin(),(x).rend()
#define tr(c,i) for(auto i :c)
#define pii pair< int,int >
#define Fast_IO() ios_base::sync_with_stdio(false);cin.tie(0)
#define pq priority_queue< pair<ll,pii> ,vector<pair<ll,pii>>,greater <pair<ll,pii>> >p;//container adapter makes ascending q
#define er(x) cout << x << " "
#define err(x,y) cout << x << " "<<y
#define fix( x ) cout << fixed << setprecision( x ) ;
const int MOD = 1000 * 1000 * 1000 + 7; // odd modulus
const int eps = 1e-9;
using namespace std;
clock_t time_p=clock();
void dem()
{
    time_p=clock()-time_p;
    cerr<<"Time Taken : "<<(float)(time_p)/CLOCKS_PER_SEC<<"\n";
}
int main()
{
    Fast_IO();
     int  n , d ;
     cin >> n >> d ;
     int cnt = 0 ;
     int i ;
     int mx = 0 ;
     for (  i = d + 1 ; i < n ; i += 2 *d + 1 )
     {
         cnt ++ ;
         mx = max ( i , mx );
     }

     cnt += mx + d <= n ;

     cout << cnt ;
}