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;
template <typename T, typename U>
void amin(T& a, U b) {
  if (a > b) a = b;
}
template <typename T, typename U>
void amax(T& a, U b) {
  if (a < b) a = b;
}
const long long INF = 1e18;
const int32_t M = 1e9 + 7;
void solve() {
  long long n;
  cin >> n;
  long long k;
  cin >> k;
  long long ans = INF, ind = 1;
  for (long long i = 1; i < k + 2; i++) {
    long long start = i;
    long long c = 1;
    while (start + 2 * k + 1 <= n) {
      c++;
      start += 2 * k + 1;
    }
    if (n - start > k) continue;
    if (ans > c) {
      ans = c;
      ind = i;
    }
  }
  cout << ans << "\n";
  while (ind <= n) {
    cout << ind << " ";
    ind += 2 * k + 1;
  }
}
void input() {}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  input();
  int i = 1;
  { solve(); }
}