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;
double PI = acos(-1);
int const N = 3e6 + 10, M = 5e6 + 10, oo = 1e9;
const int mod = oo + 7;
const long long OO = 1e18;
int cnt[3001], at[3001];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  vector<int> v;
  cin >> n >> k;
  if (k >= n) {
    cout << 1 << '\n' << 1;
    return 0;
  }
  if (k >= n / 2) {
    cout << 1 << '\n' << (n + 1) / 2 << endl;
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    i += k;
    if (i > n) break;
    at[i] = 1;
    cnt[i - k]++;
    cnt[i + k + 1]--;
    i += k;
  }
  for (int i = 1; i <= n; i++) cnt[i] += cnt[i - 1];
  if (cnt[n]) {
    for (int i = 1; i <= n; i++)
      if (at[i]) v.push_back(i);
    cout << v.size() << endl;
    for (auto e : v) cout << e << ' ';
    return 0;
  }
  v.push_back(n);
  int sub = 0;
  for (int i = n; i >= n - k; i--) {
    cnt[i]++;
    if (cnt[i] > 1) sub++;
  }
  for (int i = n; i > 0; i--)
    if (at[i])
      if (i - sub >= 1) v.push_back(i - sub);
  reverse(v.begin(), v.end());
  cout << v.size() << endl;
  for (auto e : v) cout << e << ' ';
  return 0;
}