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;
map<int, int> memo;
map<int, int> trace;
void retrace(int n, int k, int skip) {
  cout << skip + trace[n] << " ";
  if (n > 2 * k + 1) {
    retrace(n - trace[n] - k, k, skip + trace[n] + k);
  }
}
int solve(int n, int k) {
  if (n < 2 * k + 2) {
    int skewer;
    for (int i = 1; i < k + 2; i++) {
      if (k + i == n) {
        skewer = n - i + 1;
        break;
      }
    }
    trace[n] = skewer;
    return 1;
  } else {
    int ans = (int)1e9;
    int skewer;
    for (int i = k + 1; i < 2 * k + 2; i++) {
      if (n - i < k + 1) {
        continue;
      }
      if (!memo.count(n - i)) {
        memo[n - i] = solve(n - i, k);
      }
      if (trace[n - i] < k + 1) {
        continue;
      }
      ans = min(ans, memo[n - i]);
      if (ans == memo[n - i]) {
        skewer = i - k;
      }
    }
    trace[n] = skewer;
    return 1 + ans;
  }
}
int main() {
  int n, k;
  cin >> n >> k;
  if (k == 0) {
    cout << n << "\n";
    for (int i = 0; i < n; i++) {
      cout << i + 1 << " ";
    }
    cout << "\n";
  } else if (n < k + 1) {
    cout << 1 << "\n";
    cout << 1 << "\n";
  } else {
    cout << solve(n, k) << "\n";
    retrace(n, k, 0);
    cout << "\n";
  }
  return 0;
}