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;
const int N = 1e5 + 1;
int accept() {
  int n, k;
  cin >> n >> k;
  int real = n;
  int x = 2 * k + 1;
  int del = n / x;
  if (n % x == 0) {
    cout << n / x << "\n";
    int id = k + 1;
    for (int i = 1; i <= (n / x); i++) {
      cout << id << "\n";
      id += x;
    }
  } else {
    del -= 1;
    if (del < 0) {
      cout << 1 << "\n";
      cout << (n - (n / 2)) << "\n";
      return 0;
    }
    n -= (del * x);
    int need1 = 0, need2 = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = n; j >= 1; j--) {
        if ((i - (k + 1) <= 0) && (i + k) != (j - k) &&
            (i + k + 1) == (j - k) && (j + k >= n)) {
          need1 = i, need2 = j;
          int ok = 0;
          if (need1 == need2) {
            ok = 1;
            cout << del + 1 << "\n";
          } else
            cout << (del + 2) << "\n";
          cout << need1 << " ";
          need1 += k;
          need1 += (k + 1);
          for (int r = 1; r <= del; r++) {
            cout << need1 << " ";
            need1 += (2 * k + 1);
          }
          n -= need2;
          if (ok == 0) cout << real - n << " ";
          cout << "\n";
          return 0;
        }
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    accept();
  }
  return 0;
}