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 = 1009;
int n, k;
int ans = 0;
void turn(int num, int k) {
  int max_range = k + k + 1;
  int min_range = k + 1;
  if (num <= max_range) {
    cout << 1 << endl;
    cout << (num < min_range ? 1 : (num - k)) << endl;
    return;
  }
  if (num <= 2 * max_range) {
    cout << 2 << endl;
    for (int i = min_range; i <= max_range; ++i) {
      if (num - i >= min_range && num - i <= max_range) {
        cout << i - k << ' ' << num - (num - i - min_range) << endl;
        return;
      }
    }
  }
  int cnt = num / max_range;
  int re = num % max_range;
  if (re == 0) {
    cout << cnt << endl;
    for (int cur = k + 1; cur <= num; cur += k + k + 1) cout << cur << ' ';
    cout << endl;
    return;
  } else if (re < min_range) {
    re += max_range;
    ans = 2 + cnt - 1;
    cout << ans << endl;
    for (int i = min_range; i <= max_range; ++i) {
      if (re - i >= min_range && re - i <= max_range) {
        cout << i - k;
        int cur = i + k + 1;
        for (int j = 1; j < cnt; ++j, cur += k + k + 1) cout << ' ' << cur;
        cout << ' ' << num - (re - i - min_range) << endl;
        return;
      }
    }
  }
  if (re < max_range) {
    ans = cnt + 1;
    cout << ans << endl;
    cout << re - k;
    int cur = re + k + 1;
    while (cnt--) {
      cout << ' ' << cur;
      cur += k + k + 1;
    }
  }
}
int main(int argc, char const *argv[]) {
  cin >> n >> k;
  if (k == 0) {
    cout << n << endl;
    for (int i = 1; i <= n; ++i) cout << i << ' ';
    return 0;
  }
  turn(n, k);
  return 0;
}