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 read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = 10 * x + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int gcd(int x, int y) { return y ? gcd(y, x % y) : x; }
int n, k;
int a[2000];
int ans[2000005];
int main() {
  n = read();
  k = read();
  int last = n % (2 * k + 1);
  int cnt = 0;
  if (last > k + 1) {
    ans[++cnt] = n - (last - k) + 1;
    int now = k + 1;
    for (int i = 1; i <= n / (2 * k + 1); i++) {
      ans[++cnt] = now;
      now += 2 * k + 1;
    }
  } else if (last == n) {
    cout << 1 << endl;
    cout << 1 + (n / 2) << endl;
    return 0;
  } else if (last != 0) {
    int la = last / 2 + k + 1, lb = last - (last / 2) + k;
    ans[++cnt] = 1 + (last / 2);
    ans[++cnt] = n - last + (last / 2) + 1;
    int now = la + 1 + k;
    for (int i = 1; i <= (n / (2 * k + 1)) - 1; i++) {
      ans[++cnt] = now;
      now += 2 * k + 1;
    }
  } else {
    int now = k + 1;
    for (int i = 1; i <= (n / (2 * k + 1)); i++) {
      ans[++cnt] = now;
      now += 2 * k + 1;
    }
  }
  cout << cnt << endl;
  for (int i = 1; i <= cnt; i++) printf("%d ", ans[i]);
  return 0;
}