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: Accepted)

#include "bits/stdc++.h"
using namespace std;
#define MOD 1000000007
//#define MOD 998244353
const double EPS = 1e-9;
#define INF (1LL<<60)
#define D double
#define fs first
#define sc second
#define int long long
#define FOR(i,a,b) for(int i=(int)(a);i<(int)(b);++i)
#define RFOR(i,a,b) for(int i = (int)(b-1);i>=(int)(a);--i)
#define REP(i,n)  FOR(i,0,(n))
#define RREP(i,n) RFOR(i,0,(n))
#define ITR(itr,mp) for(auto itr = (mp).begin(); itr != (mp).end(); ++itr)
#define RITR(itr,mp) for(auto itr = (mp).rbegin(); itr != (mp).rend(); ++itr)
#define range(i,a,b) ((a)<=(i) && (i)<(b))
#define debug(x)  cout << #x << " = " << (x) << endl;
#define SP << " " << 
typedef pair<int,int> P;
typedef vector<int> vec;
typedef vector<vector<int>> mat;

signed main(){
  ios::sync_with_stdio(false);
  cout.tie(0);
  cout.precision(20);

  int n,d;
  cin >> n >> d;

  cout << ceil((D)n / (2 * d + 1)) << endl;

  return 0;
}