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;
template <class T>
inline bool isLeap(T y) {
  return (y % 400 == 0) || (y % 100 ? y % 4 == 0 : false);
}
template <class T>
inline T gcd(T a, T b) {
  return (b) == 0 ? (a) : gcd((b), ((a) % (b)));
}
template <class T>
inline T lcm(T a, T b) {
  return ((a) / gcd((a), (b)) * (b));
}
template <class T>
inline T BigMod(T Base, T power, T M = 1000000007) {
  if (power == 0) return 1;
  if (power & 1)
    return ((Base % M) * (BigMod(Base, power - 1, M) % M)) % M;
  else {
    T y = BigMod(Base, power / 2, M) % M;
    return (y * y) % M;
  }
}
template <class T>
inline T ModInv(T A, T M = 1000000007) {
  return BigMod(A, M - 2, M);
}
int fx[] = {-1, +0, +1, +0, +1, +1, -1, -1, +0};
int fy[] = {+0, -1, +0, +1, +1, -1, +1, -1, +0};
int day[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> ans;
  int mn = INT_MAX;
  for (int i = 1; i <= 1 + k; i++) {
    int pk = i;
    int cnt = 1;
    vector<int> v;
    while (pk <= n) {
      cnt++;
      if (pk + k + 1 > n) {
        if (cnt < mn) {
          mn = cnt;
          v.push_back(pk);
          ans = v;
        }
        break;
      } else if (pk + k + 1 <= n) {
        if (pk + k + 1 + k <= n) {
          v.push_back(pk);
          pk = pk + k + 1 + k;
        } else
          break;
      }
    }
  }
  cout << ans.size() << endl;
  for (auto i : ans) cout << i << " ";
  return 0;
}