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 dx8[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy8[] = {1, -1, 1, -1, 0, 0, -1, 1};
int dx4[] = {0, 0, 1, 1};
int dy4[] = {1, -1, 1, -1};
int arr[1005][1005];
int main() {
  int n, k, j = -1, idx, x, rem;
  cin >> n >> k;
  int a[n + 1];
  for (int i = 1; i <= n; i++) a[i] = i;
  vector<int> v;
  x = (2 * k) + 1;
  rem = n % x;
  if (rem == 0 || rem > k) {
    for (int i = x; i <= n; i += x) {
      idx = i - k;
      v.push_back(a[idx]);
      j = i;
    }
    if (rem >= k + 1) v.push_back(a[j + k + 1]);
  } else if (rem <= k) {
    v.push_back(a[1]);
    j = (k + 1);
    for (int i = x + k + 1; i <= n; i += x) {
      idx = i - k;
      v.push_back(a[idx]);
      j = i;
    }
    if (n > k) v.push_back(a[j + k + 1]);
  }
  cout << v.size() << endl;
  for (int i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << endl;
  return 0;
}