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;
using ll = long long;
using ull = unsigned long long;
const long long MOD = 1e9 + 7;
const long long INF = 5e18;
const double DINF = 5e14;
const double eps = 1e-10;
const int di[4] = {1, 0, -1, 0}, dj[4] = {0, 1, 0, -1};
int n, k;
ll dp[3010];
int back[3010], it[3010];
vector<int> ans;
int main() {
  cin >> n >> k;
  for (int i = 0; i < n + 1; i++) {
    dp[i] = INF;
  }
  for (int i = 0; i <= k; i++) {
    dp[min(i + k, n - 1)] = 1;
    back[min(n - 1, i + k)] = i + 1;
    it[min(n - 1, i + k)] = i + 1;
    if (dp[n - 1] == 1) {
      cout << 1 << endl;
      cout << i + 1 << endl;
      return 0;
    }
  }
  for (int i = k; i < n; i++) {
    if (dp[min(n - 1, i + k + 1 + k)] > dp[i] + 1) {
      dp[min(n - 1, i + k + 1 + k)] = dp[i] + 1;
      back[min(n - 1, i + k + 1 + k)] = i + 1;
      it[min(n - 1, i + k + 1 + k)] = min(n, i + k + 2);
    }
  }
  cout << dp[n - 1] << endl;
  int cnt = dp[n - 1], now = n - 1;
  while (cnt) {
    ans.emplace_back(it[now]);
    now = back[now] - 1;
    cnt--;
  }
  for (int i = dp[n - 1] - 1; i >= 0; i--) {
    if (i != 0)
      cout << ans[i] << ' ';
    else
      cout << ans[i] << endl;
  }
  return 0;
}