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 int N = 2e5 + 5;
const double eps = 1e-9;
const int MAX_N = 1e6 + 5;
const int inf = 2e9;
const long long Inf = (long long)1e18;
const long long M = (long long)1e9 + 7;
int Int() {
  int x;
  scanf("%d", &x);
  return x;
}
long long Long() {
  long long x;
  scanf("%lld", &x);
  return x;
}
void Char(char* ch) { scanf("%s", ch); }
template <typename t>
inline t abs(t a) {
  if (a >= 0) return a;
  return -a;
}
template <typename t>
inline t gcd(t a, t b) {
  return !b ? a : gcd(b, a % b);
}
template <typename t>
inline t lcm(t a, t b) {
  return a * (b / gcd(a, b));
}
template <typename T>
inline T modpow(T b, T p, T Mod) {
  T res = 1;
  while (p > 0) {
    if (p & 1) {
      res *= b;
      res %= Mod;
    }
    b *= b;
    b %= Mod;
    p >>= 1;
  }
  return res;
}
template <typename T>
inline T ModInv(T b, T Mod) {
  return modpow(b, Mod - 2, Mod);
}
template <typename T>
inline T pwr(T b, T p) {
  T res = 1;
  while (p > 0) {
    if (p & 1) {
      res *= b;
    }
    b *= b;
    p >>= 1;
  }
  return res;
}
int e1, e2, d1;
template <typename T>
inline void extgcd(T a, T b) {
  if (!b) {
    e1 = 1, e2 = 0, d1 = a;
    return;
  }
  extgcd(b, a % b);
  T x1 = e2;
  T y1 = e1 - (a / b) * e2;
  e1 = x1;
  e2 = y1;
}
template <typename T>
inline vector<T> Unique(vector<T> v) {
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  return v;
}
vector<string> split(const string& s, char c) {
  vector<string> v;
  stringstream ss(s);
  string x;
  while (getline(ss, x, c)) v.emplace_back(x);
  return move(v);
}
void err(vector<string>::iterator it) {}
template <typename T, typename... Args>
void err(vector<string>::iterator it, T a, Args... args) {
  cout << it->substr((*it)[0] == ' ', it->length()) << " = " << a << " ";
  err(++it, args...);
}
int main() {
  int t = 1, tc = 0;
  while (t--) {
    int n = Int(), k = Int();
    int f = (k << 1) + 1;
    if (n == 1) {
      puts("1\n1");
    } else if (f >= n) {
      puts("1");
      int d = min(n, k + 1);
      printf("%d\n", d);
    } else if (n % f == 0) {
      cout << n / f << '\n';
      for (int i = 1; i <= n; i += f) {
        printf("%d ", i + k);
      }
    } else {
      std::vector<int> v;
      for (int i = k + 1; i <= n; i += f) {
        v.push_back(i);
      }
      if (v.back() < n) {
        if (n % f <= k) {
          int l = k - (n % f) + 1;
          for (int i = 0; i < (int)v.size(); i++) {
            v[i] -= l;
          }
          v.push_back(n);
        } else {
          int d = (n % f) - (k + 1);
          v.push_back(n - d);
        }
      }
      v = Unique(v);
      printf("%d\n", (int)v.size());
      for (int i : v) cout << i << ' ';
    }
    puts("");
  }
  return 0;
}