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>
#include <set>
#define REP(i, n) for(int i = 0;i < n;i++)
#define REPR(i, n) for(int i = n;i >= 0;i--)
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define FORR(i, m, n) for(int i = m;i >= n;i--)
#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
#define llong long long
#define pb(a) push_back(a)

#define INF 100000000
#define MOD 100000007

#define writeln(n) cout<<n<<endl

using namespace std;
typedef pair<int, int> P;
typedef pair<llong, llong> LP;
typedef pair<int, P> PP;
typedef pair<llong, LP> LPP;
//vector<tuple<long long, long long, long long>>vec;

template<class T>
inline bool chmax(T &a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

template<class T>
inline bool chmin(T &a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}

llong pownew(llong m, llong n) {
    if (n == 0) { return 1; }
    if (n % 2 == 0) { return pow(m*m, n/2); }
    return pow(m, n-1) * m;
}

bool isprime(int p) {
    if (p == 1) return false;
    for (int i = 2; i < p; i++) {
        if (p%i == 0) return false;
    }
    return true;
}

int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};

const int inf=100000000;

long long modpow(long long a, long long n, long long mod) //a^n
{
    long long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}

llong gcd(llong a, llong b) {
    return b != 0 ? gcd(b, a % b) : a;
}
//setprecision(10);



int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n,d; cin>>n>>d;
    int ans=0;
    if(n<=d) ans=1;
    else {

        if (n % (2 * d + 1) != 0) ans = n / (2 * d + 1) + 1;
        else ans = n / (2 * d + 1);
    }
    cout<<ans<<endl;
    return  0;
}