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 long long mod = 1000000007;
const long long inf = LLONG_MAX;
const long long ninf = LLONG_MIN;
void primeFactors(long long p) {
  while (!(p % 2)) {
    p /= 2;
  }
  for (long long i = 3; i <= sqrt(p); i += 2) {
    while (!(p % i)) {
      p /= i;
    }
  }
  if (p > 2) {
    ;
  }
}
template <typename... T>
void input(T &...args) {
  ((cin >> args), ...);
}
template <typename... T>
void print(T &&...args) {
  ((cout << args << " "), ...);
}
template <typename... T>
void deb(T &&...args) {
  ((cerr << args << " "), ...);
  cerr << "\n";
}
void forv(vector<long long> &v, long long n) {
  long long x;
  for (long long i = 0; i < n; i++) {
    cin >> x;
    v.push_back(x);
  }
}
void printv(vector<long long> v) {
  for (auto i : v) {
    cout << i << " ";
  }
}
void fora(long long *a, long long n) {
  for (long long i = 0; i < n; i++) cin >> a[i];
}
void printa(long long *a, long long n) {
  for (long long i = 0; i < n; i++) cout << a[i] << " ";
}
void dmap(map<long long, long long> mp) {
  for (auto i : mp) cout << i.first << " " << i.second << "\n";
}
void dmmap(multimap<long long, long long> mp) {
  for (auto i : mp) cout << i.first << " " << i.second << "\n";
}
void sortra(long long *a, long long n) { sort(a, a + n, greater<long long>()); }
void sortrv(vector<long long> &v) {
  sort(v.begin(), v.end(), greater<long long>());
}
inline void guts() {}
void solve() {
  long long n, k;
  cin >> n >> k;
  if (2 * k + 1 >= n) {
    cout << 1 << "\n" << n / 2 + 1;
    return;
  }
  long long m = 2 * k + 1;
  if (n % m == 0) {
    cout << n / m << "\n";
    long long dd = 2 * k + 1;
    for (long long i = k + 1; i <= n; i += dd) cout << i << " ";
    return;
  }
  long long kk = n / m;
  if (k + 1 <= n % m) {
    cout << kk + 1 << "\n";
    for (long long i = k + 1; i <= n; i += m) cout << i << " ";
    return;
  }
  long long dd = 2 * k + 1;
  cout << n / m + 1 << "\n";
  for (long long i = n % m; i <= n; i += dd) cout << i << " ";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  guts();
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}