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 (GCC 5.4.1) (judge verdict: Accepted)

#define _CRT_SECURE_NO_WARNINGS

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>

/* define */
#define Yes(X) puts ( (X) ? "Yes" : "No")
#define YES(X) puts ( (X) ? "YES" : "NO")
#define FOR(i, a, b) for ( int i = (a); i < (b) ; i++)
#define REP(i, n) for ( int i = 0; i < (n) ; i++)
#define max(a, b) (((a) > (b)) ? (a) : (b))
#define min(a, b) (((a) < (b)) ? (a) : (b))
#define max3(a, b, c) ((max((a), (b)) > (c)) ? max((a), (b)) : (c))
#define min3(a, b, c) ((min((a), (b)) < (c)) ? min((a), (b)) : (c))

/* const */
const int MOD = 1000000007;

/* alias */
typedef long long ll;
typedef unsigned  u;

/* function */
int cmp(const void *a, const void *b) {
	return *(int*)a - *(int*)b;
}
int cmp2(const void *a, const void *b) {
	return *(int*)b - *(int*)a;
}
int gcd(int a, int b) {
	if (b != 0)return gcd(b, a % b);
	return a;
}
ll Lgcd(ll a, ll b) {
	if (b != 0) return Lgcd(b, a % b);
	return a;
}
int lcm(int a, int b) {
	return a / gcd(a, b) * b;
}
ll Llcm(ll a, ll b) {
	return a / Lgcd(a, b) * b;
}

/* main */

int main() {
	int n, d, r;
	scanf("%d%d", &n, &d);
	if (n % (d * 2 + 1)) r = n / (2 * d + 1) + 1;
	else r = n / (2 * d + 1);
	printf("%d", r);
	return 0;
}