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 main345() {
  int a[10];
  int b[5] = {0, 5, 3, 1};
  while (scanf("%d %d %d %d %d %d", &a[1], &a[2], &a[3], &a[4], &a[5], &a[6])) {
    if (a[1] == 0 && a[2] == 0 && a[3] == 0 && a[4] == 0 && a[5] == 0 &&
        a[6] == 0)
      break;
    int sum = 0;
    sum += a[6];
    sum += a[5];
    sum += a[4];
    sum += (a[3] + 3) / 4;
    int cnt = 0;
    cnt += a[4] * 5 + b[a[3] % 4];
    if (cnt < a[2]) sum += (a[2] - cnt + 8) / 9;
    cnt = sum * 36 - a[6] * 36 - a[5] * 25 - a[4] * 16 - a[3] * 9 - a[2] * 4;
    if (cnt < a[1]) sum += (a[1] - cnt + 35) / 36;
    printf("%d\n", sum);
  }
  return 0;
}
int main() {
  int n, k;
  while (cin >> n >> k) {
    if (n <= k + k + 1) {
      cout << 1 << endl;
      cout << n / 2 + 1 << endl;
    } else {
      int ans = 1;
      int u = k + k + 2;
      while (ans++) {
        if (u > n) break;
        u += k + k + 1;
      }
      u -= k + k + 1;
      int a = 1, b = n;
      cout << ans - 1 << endl;
      for (int i = 0; i < n - u; i++)
        if (i % 2)
          a++;
        else
          b--;
      while (a != b) {
        cout << a << " ";
        a += k + k + 1;
      }
      cout << b << endl;
    }
  }
}