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;
const double pi = acos(-1.0);
void fast() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long int comb(long long int m, long long int n) {
  long long int mat[100][100];
  long long int i, j;
  if (n > m) return 0;
  if ((n == 0) || (m == n)) return 1;
  for (j = 0; j < n; j++) {
    mat[0][j] = 1;
    if (j == 0) {
      for (i = 1; i <= m - n; i++) mat[i][j] = i + 1;
    } else {
      for (i = 1; i <= m - n; i++) mat[i][j] = mat[i - 1][j] + mat[i][j - 1];
    }
  }
  return (mat[m - n][n - 1]);
}
int main() {
  long long int n, m;
  cin >> n >> m;
  long long int x = 0;
  long long int i = 0;
  while (x < n) {
    i++;
    x = x + 2 * m + 1;
  }
  x = x - n;
  long long int s = x;
  cout << i << '\n';
  if (x <= m) {
    s = m - x + 1;
    for (int j = 0; j < i; j++) {
      cout << s << " ";
      s = s + 2 * m + 1;
    }
  } else {
    s = x - m;
    s = m - s + 1;
    for (int j = 0; j < i; j++) {
      cout << s << " ";
      s = s + 2 * m + 1;
    }
  }
}