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;
unordered_map<long long, long long> hm;
void countFreq(long long a[], long long n) {
  for (long long i = 0; i < n; i++) hm[a[i]]++;
  unordered_set<int> s;
  for (long long i = 0; i < n; i++) {
    if (s.find(a[i]) == s.end()) {
      s.insert(a[i]);
    }
  }
}
int main() {
  long long t = 1;
  while (t--) {
    long long n, k, ans = 1000000000;
    cin >> n >> k;
    if (n <= k + 1) {
      cout << 1 << endl;
      cout << 1 << endl;
      return 0;
    }
    for (long long i = 1; i <= k + 1; i++) {
      long long cnt = 1;
      long long j = n - (i + k);
      cnt += (j / (2 * k + 1));
      long long m = (j % (2 * k + 1));
      if (m == 0) {
        if (cnt < ans) ans = cnt;
      }
      if (m >= k + 1 && m <= 2 * k + 1) {
        cnt++;
        if (cnt < ans) ans = cnt;
      }
    }
    for (long long i = 1; i <= k + 1; i++) {
      long long cnt = 1;
      long long j = n - (i + k);
      cnt += (j / (2 * k + 1));
      long long m = (j % (2 * k + 1));
      if (m == 0) {
        if (cnt == ans) {
          cout << cnt << endl;
          cout << i << " ";
          for (long long s = 2 * k + i + 1; s <= n; s += 2 * k + 1) {
            cout << s << " ";
          }
          return 0;
        }
      }
      if (m >= k + 1 && m <= 2 * k + 1) {
        cnt++;
        if (cnt == ans) {
          cout << cnt << endl;
          cout << i << " ";
          for (long long s = 2 * k + i + 1; s <= n; s += 2 * k + 1) {
            cout << s << " ";
          }
          return 0;
        }
      }
    }
  }
  return 0;
}