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;
int main() {
  int N, K;
  scanf("%d%d", &N, &K);
  if (K >= N) {
    printf("1\n1\n");
    return 0;
  }
  if (2 * K + 1 >= N) {
    printf("1\n%d\n", K + 1);
    return 0;
  }
  if (N % (2 * K + 1) == 0) {
    printf("%d\n", N / (2 * K + 1));
    int curr = K + 1;
    bool was = false;
    while (curr <= N) {
      if (was) {
        printf(" ");
      }
      printf("%d", curr);
      curr += 2 * K + 1;
      was = true;
    }
  } else {
    int sz = N / (2 * K + 1) + 1, last = 1;
    printf("%d\n", sz);
    int curr = 1;
    bool was = false;
    vector<int> ans;
    while (curr <= N) {
      ans.push_back(curr);
      last = curr + K + 1;
      curr += 2 * K + 1;
    }
    if (last <= N) {
      for (int trying = 2; trying <= N && last <= N; trying++) {
        curr = trying, last = 1;
        ans.clear();
        while (curr <= N) {
          ans.push_back(curr);
          last = curr + K + 1;
          curr += 2 * K + 1;
        }
      }
      for (int i = 0; i < sz; i++) {
        if (i > 0) {
          printf(" ");
        }
        printf("%d", ans[i]);
      }
      return 0;
    } else {
      for (int i = 0; i < sz; i++) {
        if (i > 0) {
          printf(" ");
        }
        printf("%d", ans[i]);
      }
    }
  }
  printf("\n");
  return 0;
}