Problem: Long story short, shashlik is Miroslav's favorite food. Shashlik is prepared on several skewers simultaneously. There are two states for each skewer: initial and turned over.

This time Miroslav laid out n skewers parallel to each other, and enumerated them with consecutive integers from 1 to n in order from left to right. For better cooking, he puts them quite close to each other, so when he turns skewer number i, it leads to turning k closest skewers from each side of the skewer i, that is, skewers number i - k, i - k + 1, ..., i - 1, i + 1, ..., i + k - 1, i + k (if they exist). 

For example, let n = 6 and k = 1. When Miroslav turns skewer number 3, then skewers with numbers 2, 3, and 4 will come up turned over. If after that he turns skewer number 1, then skewers number 1, 3, and 4 will be turned over, while skewer number 2 will be in the initial position (because it is turned again).

As we said before, the art of cooking requires perfect timing, so Miroslav wants to turn over all n skewers with the minimal possible number of actions. For example, for the above example n = 6 and k = 1, two turnings are sufficient: he can turn over skewers number 2 and 5.

Help Miroslav turn over all n skewers.

Input

The first line contains two integers n and k (1 ≤ n ≤ 1000, 0 ≤ k ≤ 1000) — the number of skewers and the number of skewers from each side that are turned in one step.

Output

The first line should contain integer l — the minimum number of actions needed by Miroslav to turn over all n skewers. After than print l integers from 1 to n denoting the number of the skewer that is to be turned over at the corresponding step.

Examples

Input

7 2


Output

2
1 6 


Input

5 1


Output

2
1 4 

Note

In the first example the first operation turns over skewers 1, 2 and 3, the second operation turns over skewers 4, 5, 6 and 7.

In the second example it is also correct to turn over skewers 2 and 5, but turning skewers 2 and 4, or 1 and 5 are incorrect solutions because the skewer 3 is in the initial state after these operations.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int rnk[150001];
int parnt[150001];
bool prim[100001];
void seive() {
  prim[0] = prim[1] = 1;
  for (unsigned long long i = 2; i < 100001; i++) {
    if (prim[i] == 0) {
      for (unsigned long long j = i * i; j < 100001; j += i) prim[j] = 1;
    }
  }
}
int findparnt(int x) {
  if (parnt[x] == x) return x;
  return parnt[x] = findparnt(parnt[x]);
}
bool iscon(int x, int y) { return (findparnt(x) == findparnt(y)); }
void conn(int x, int y) {
  int px = findparnt(x);
  int py = findparnt(y);
  if (px == py) return;
  if (rnk[px] > rnk[py]) {
    parnt[py] = px;
  } else if (rnk[px] < rnk[py]) {
    parnt[px] = py;
  } else {
    rnk[px]++;
    parnt[py] = px;
  }
}
long long gcd(long long x, long long y) {
  if (x % y == 0) return y;
  return gcd(y, x % y);
}
unsigned long long fastpow(unsigned long long b, unsigned long long p) {
  if (p == 0) return 1;
  unsigned long long t = fastpow(b, p / 2);
  if (p % 2 == 0) return t * t;
  return t * t * b;
}
unsigned long long lcm(unsigned long long a, unsigned long long b) {
  return ((a * b) / gcd(a, b));
}
double dis(double x, double y, double x1, double y1) {
  return sqrt((x - x1) * (x - x1) + (y - y1) * (y - y1));
}
bool compare(const pair<pair<int, int>, int> &a,
             const pair<pair<int, int>, int> &b) {
  if (a.first.second > b.first.second) return 1;
  if (a.first.second == b.first.second && a.second < b.second) return 1;
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  vector<int> v, v1;
  int n, y;
  cin >> n >> y;
  for (int i = 1; i < n + 1; i += 2 * y + 1) {
    v.push_back(i);
    if (i + (2 * y) + 1 > n && i + y < n) {
      v.push_back(n);
      break;
    }
  }
  for (int i = y + 1; i < n + 1; i += 2 * y + 1) {
    v1.push_back(i);
    if (i + (2 * y) + 1 > n && i + y < n) {
      v1.push_back(n);
      break;
    }
  }
  if (v.size() > v1.size() && v1.size() != 0) {
    cout << v1.size() << "\n";
    for (long long(i) = (0); (i) < (v1.size()); ++(i)) cout << v1[i] << " ";
  } else {
    cout << v.size() << "\n";
    for (long long(i) = (0); (i) < (v.size()); ++(i)) cout << v[i] << " ";
  }
}