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;
vector<int> answer;
int m, M;
int main() {
  int n, k;
  cin >> n >> k;
  m = k + 1;
  M = 2 * k + 1;
  int put = 0;
  bool odd = true;
  if (n <= M) {
    printf("%d\n%d", 1, n / 2 + 1);
    return 0;
  }
  if (n % M == 0) {
    odd = false;
  }
  if (!odd) {
    put = k + 1;
    printf("%d\n", n / M);
    for (int i = 0; i < n / M; i++) {
      printf("%d ", put + i * M);
    }
  } else {
    int s = n / M + 1;
    int left = n - M * (s - 2);
    int a, b;
    for (a = m; a < M; a++) {
      for (b = a; b < M; b++) {
        if (a + b == left) break;
      }
      if (a + b == left) break;
    }
    put = a - k;
    answer.push_back(put);
    for (int i = 1; i <= s - 2; i++) {
      answer.push_back(put + M * i);
    }
    answer.push_back(n - (b - k - 1));
    printf("%d\n", answer.size());
    for (int i = 0; i < answer.size(); i++) {
      printf("%d ", answer[i]);
    }
  }
  return 0;
}