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 MAXN = -1;
int n, k;
int l[3000];
int main() {
  while (scanf("%d%d", &n, &k) == 2) {
    if (n < 2 * k + 1) {
      printf("1\n");
      printf("%d\n", n / 2 + 1);
      continue;
    }
    vector<int> ans;
    int ansz = 10000;
    for (int i = (0); i < (int)(k + 1); i++) {
      vector<int> tans;
      memset(l, 0, sizeof(l));
      for (int j = (0); j < (int)(i); j++) l[j] = !l[j];
      for (int j = (i); j < (int)(i + k + 1); j++) l[j] = !l[j];
      tans.push_back(i);
      int p = i + k + k + 1;
      while (p < n) {
        for (int j = (0); j < (int)(k); j++)
          if (p - j - 1 >= 0) l[p - j - 1] = !l[p - j - 1];
        for (int j = (p); j < (int)(p + k + 1); j++)
          if (j < n) l[j] = !l[j];
        tans.push_back(p);
        p = p + k + k + 1;
      }
      if (l[n - 1] == 1) {
        if (ansz > (int)tans.size()) {
          ans = tans;
          ansz = ans.size();
        }
      }
    }
    sort(ans.begin(), ans.end());
    printf("%d\n", (int)ans.size());
    for (int i = (0); i < (int)(ans.size()); i++) printf("%d ", ans[i] + 1);
    puts("");
  }
  return 0;
}