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)

//------------------------------------------
// C++ templete 
//------------------------------------------
#include <bits/stdc++.h>
#include <iomanip>
using namespace std;
using ll = long long;

#define endl "\n"

//typedef
//------------------------------------------
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef vector<PII> VP;

//REPEAT
//------------------------------------------
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)

//container util
//------------------------------------------
#define pb push_back
#define paired make_pair
#define ALL(a) (a).begin(),(a).end()
#define PRINT(V) for (auto v : (V)) cout << v << " "
#define SORT(V) sort((V).begin(),(V).end())
#define RSORT(V) sort((V).rbegin(), (V).rend())

//constant
//------------------------------------------
const int MOD = 1000000007;
const int INF = 1061109567;
const double EPS = 1e-10;
const double PI  = acos(-1.0);
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};

//math
//------------------------------------------
int QP(int a,int b){int ans=1;do{if(b&1)ans=1ll*ans*a%MOD;a=1ll*a*a%MOD;}while(b>>=1);return ans;}
int QP(int a,int b,int MOD){int ans=1;do{if(b&1)ans=1ll*ans*a%MOD;a=1ll*a*a%MOD;}while(b>>=1);return ans;}
int GCD(int a,int b){return b?GCD(b,a%b):a;}

//debug
//------------------------------------------
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;

//動的計画法で役に立つ関数
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

//最大公約数と最小公倍数
ll gcd(ll a, ll b) {
    if(b==0)return a;
    return gcd(b, a%b);
}

ll lcm(ll a, ll b) {
  ll g = gcd(a, b);
  return a/g*b;
}

vector<vector<long long> > comb(int n) {
  vector<vector<long long> > v(n + 1,vector<long long>(n + 1, 0));
  for (int i = 0; i < v.size(); i++) {
    v[i][0] = 1;
    v[i][i] = 1;
  }
  for (int j = 1; j < v.size(); j++) {
    for (int k = 1; k < j; k++) {
      v[j][k] = (v[j - 1][k - 1] + v[j - 1][k]);
    }
  }
  return v;
}

int main() 
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
  
    double N, D; cin >> N >> D;
    cout << ceil(N/(2*D+1)) << endl;

    return 0;
}