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;
template <class T>
T mceil(T a, T b) {
  if (a % b == 0)
    return a / b;
  else
    return a / b + 1;
}
template <class T>
T mfloor(T a, T b) {
  if (a % b == 0)
    return a / b;
  else
    return a / b - 1;
}
const unsigned int M = 1000000007;
bool pal(string &s) {
  if (s == string(s.rbegin(), s.rend()))
    return true;
  else
    return false;
}
long long fact(long long n) {
  long long res = 1;
  for (long long i = 2; i <= n; i++) res = res * i;
  return res;
}
int ncr(int n, int k) {
  double res = 1;
  for (int i = 1; i <= k; ++i) res = res * (n - k + i) / i;
  return (int)(res + 0.01);
}
unsigned long long factorial(long long int n) {
  unsigned long long f = 1;
  for (int i = 1; i <= n; i++) f = (f * i) % M;
  return f;
}
unsigned int cntsetbits(int x) {
  unsigned int nsb = 0;
  while (x) {
    ++nsb;
    x = x & (x - 1);
  }
  return nsb;
}
long long negmod(long long a) { return (a % M + M) % M; }
void solve() {
  int n, k;
  cin >> n >> k;
  vector<int> v;
  int j = 0;
  int shift = 0;
  for (int i = 1; i <= n; ++i) {
    if (i == 1) {
      j = min(n, k + 1);
      v.push_back(j);
      if (j == n || j + k + 1 > n) break;
    } else {
      j = j + (2 * k) + 1;
      v.push_back(j);
      if (j >= n || j + k + 1 > n) {
        shift = j - n;
        break;
      }
    }
  }
  if (shift < 0) shift = 0;
  cout << v.size() << endl;
  for (int &a : v) cout << a - shift << " ";
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
}