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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long long INF = 1000LL * 1000 * 1000 * 1000 * 1000 * 1000;
const int inf = 1000 * 1000 * 1000;
const long double PI = acos(-1.0);
const long long mod1 = inf + 7;
const long long mod2 = inf + 9;
const int MAXN = 2001;
const long double EPS = 1e-9;
int hp = 179;
int main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  srand(time(NULL));
  ;
  int n, k;
  cin >> n >> k;
  if (k == 0) {
    cout << n << '\n';
    for (int i = 1; i <= n; ++i) {
      cout << i << " ";
    }
    return 0;
  }
  if (k >= n - 1) {
    cout << 1 << "\n" << 1;
    return 0;
  }
  if (2 * k + 1 >= n) {
    cout << 1 << "\n" << (n + 1) / 2 << '\n';
    return 0;
  }
  if (n % (2 * k + 1) == 0) {
    cout << n / (2 * k + 1) << '\n';
    int lt = 0;
    while (lt < n) {
      cout << lt + k + 1 << ' ';
      lt += 2 * k + 1;
    }
    return 0;
  }
  if (n % (2 * k + 1) < k + 1) {
    vector<int> ans;
    int lt = 1 + k;
    for (;;) {
      ans.push_back(lt - k);
      if (n - lt < 2 * k + 1) {
        ans.push_back(lt + k + 1);
        break;
      }
      lt += 2 * k + 1;
    }
    cout << ans.size() << '\n';
    for (int i = 0; i < ans.size(); ++i) cout << ans[i] << " ";
    ;
    return 0;
  }
  vector<int> ans;
  int lt = 2 * k + 1;
  for (;;) {
    ans.push_back(lt - k);
    if (n - lt < 2 * k + 1) {
      ans.push_back(lt + k + 1);
      break;
    }
    lt += 2 * k + 1;
  }
  cout << ans.size() << '\n';
  for (int i = 0; i < ans.size(); ++i) cout << ans[i] << " ";
  ;
}