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 double pi =
    3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  int v = 2 * k + 1;
  if (n <= k + 1)
    printf("1\n1\n");
  else if (n % v == 0) {
    printf("%d\n", n / v);
    int ile = 0, pos = k + 1;
    while (ile < n) {
      printf("%d ", pos);
      pos += v;
      ile += v;
    }
    printf("\n");
  } else if (n % v > k) {
    int r = n % v;
    int pos = 1 + r - (k + 1);
    int ile = r;
    printf("%d\n%d ", n / v + 1, pos);
    pos = ile + k + 1;
    while (ile < n) {
      printf("%d ", pos);
      pos += v;
      ile += v;
    }
    printf("\n");
  } else {
    int res = n * n;
    for (int i = (k + 1); i <= (v); ++i)
      for (int j = (k + 1); j <= (v); ++j)
        if ((n - i - j) >= 0 && (n - i - j) % v == 0)
          res = min(res, 2 + (n - i - j) / v);
    printf("%d\n", res);
    for (int i = (k + 1); i <= (v); ++i)
      for (int j = (k + 1); j <= (v); ++j)
        if ((n - i - j) >= 0 && (n - i - j) % v == 0 &&
            2 + (n - i - j) / v == res) {
          int pos = 1 + i - (k + 1);
          int ile = i;
          printf("%d %d ", pos, n - (j - (k + 1)));
          n -= j;
          pos = ile + k + 1;
          while (ile < n) {
            printf("%d ", pos);
            pos += v;
            ile += v;
          }
          printf("\n");
          return 0;
        }
  }
  return 0;
}