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;
  cin >> n >> k;
  int l;
  vector<int> ans;
  if (k == 0) {
    cout << n << endl;
    for (int i = 1; i <= n; i++) cout << i << ' ';
    cout << endl;
    return 0;
  }
  if (n <= 2 * k + 1) {
    l = 1;
    cout << 1 << endl;
    int ans2;
    if (n > k)
      ans2 = k + 1;
    else
      ans2 = 1;
    cout << ans2 << endl;
    return 0;
  }
  if (n % (2 * k + 1) == 0) {
    l = n / (2 * k + 1);
    ans.push_back(k + 1);
    int temp = k + 1 + (2 * k + 1);
    while (temp < n) {
      ans.push_back(temp);
      temp = temp + 2 * k + 1;
    }
  } else if (n % (2 * k + 1) >= k + 1) {
    int t1 = n / (2 * k + 1);
    l = 1;
    int n2 = (2 * k + 1) * t1;
    ans.push_back(k + 1);
    int temp = k + 1 + (2 * k + 1);
    while (temp < n2) {
      l++;
      ans.push_back(temp);
      temp = temp + 2 * k + 1;
    }
    ans.push_back(n2 + 1 + k);
    l++;
  } else {
    int rem = n % (2 * k + 1);
    l = 0;
    int miss1, miss2;
    if (rem % 2 == 0) {
      miss1 = rem / 2;
      miss2 = rem / 2;
    } else {
      miss1 = rem / 2;
      miss2 = (rem / 2) + 1;
    }
    int temp = miss1 + 1;
    while (temp <= n) {
      l++;
      ans.push_back(temp);
      temp = temp + 2 * k + 1;
    }
  }
  cout << l << endl;
  for (int i = 0; i < l; i++) {
    cout << ans[i] << ' ';
  }
  cout << endl;
  return 0;
}