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 = 1e9 + 7;
const long long inf = 1e17;
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
long long ceil_div(long long a, long long b) { return (a + b - 1) / b; }
long long power(long long a, long long b) {
  if (b == 0) return 1;
  if (a == 1) return 1;
  long long x = power(a, b / 2);
  if (b % 2 == 0)
    return (x * x);
  else
    return (a * (x * x));
}
void solve() {
  long long n, k;
  cin >> n >> k;
  if (k >= n) {
    cout << 1 << '\n';
    cout << 1 << '\n';
    return;
  }
  vector<long long> v;
  for (long long i = k + 1; i <= 2 * k + 1; i++) v.push_back(i);
  long long dp[n + 1];
  for (long long i = 0; i <= n; i++) dp[i] = inf;
  dp[0] = 0;
  for (long long i = 1; i <= n; i++) {
    for (long long j = 0; j < v.size(); j++) {
      if (i - v[j] >= 0) dp[i] = min(dp[i], 1 + dp[i - v[j]]);
    }
  }
  cout << dp[n] << '\n';
  long long temp = dp[n];
  if (dp[n] == 1) {
    cout << (1 + n) / 2;
  } else {
    vector<long long> ans;
    long long x;
    x = k + 1;
    while (dp[n]) {
      --dp[n];
      ans.push_back(x);
      x += (2 * k + 1);
    }
    long long var = 0;
    if (ans[ans.size() - 1] > n) var = ans[ans.size() - 1] - n;
    for (long long x : ans) cout << x - var << " ";
  }
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout << setprecision(15) << fixed;
  long long t;
  t = 1;
  while (t--) {
    solve();
  }
}