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;
void in() { return; }
template <typename T, typename... Types>
void in(T &a, Types &...b) {
  cin >> (a);
  in(b...);
}
void o() { return; }
template <typename T, typename... Types>
void o(T a, Types... b) {
  cout << (a);
  cout << ' ';
  o(b...);
}
bool sortin(const pair<long long int, long long int> &e,
            const pair<long long int, long long int> &f) {
  return (e.first < f.first);
}
bool POT(long long int x) { return x && (!(x & (x - 1))); }
long long int i, j, k, l, m, n, p, q, r, a, b, c, x, y, z, ts, mn = 1e18,
                                                               mod = 1e9 + 7;
long long int ar[250005], br[250005], xr[250005], tem[250005];
int main() {
  {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
  };
  in(n, k);
  long long int carry = 0;
  while (n > 0) {
    n = n - k;
    ar[c++] = n;
    if (n < 1) {
      carry = 1 - n;
      break;
    }
    n = n - k - 1;
  }
  o(c);
  cout << endl;
  for (long long int i = c - 1; i >= 0; i--) o(ar[i] + carry);
}