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;
long long mult(long long a, long long b, long long p = 1000000007) {
  return ((a % p) * (b % p)) % p;
}
long long add(long long a, long long b, long long p = 1000000007) {
  return (a % p + b % p) % p;
}
long long fpow(long long n, long long k, long long p = 1000000007) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
long long inv(long long a, long long p = 1000000007) {
  return fpow(a, p - 2, p);
}
int main() {
  ios::sync_with_stdio(0);
  long long n, k;
  cin >> n >> k;
  if (k == 0) {
    cout << n << "\n";
    for (long long i = 0; i < n; i++) {
      cout << i + 1 << " ";
    }
  } else if (2 * k + 1 >= n) {
    cout << 1 << "\n" << 1 + n / 2 << "\n";
  } else {
    long long p = 2 * k + 1;
    if (n % p == 0) {
      cout << n / p << "\n";
      long long cnt = 0;
      for (long long i = 1;; i += p) {
        cout << i + k << " ";
        cnt++;
        if (cnt == n / p) {
          break;
        }
      }
    } else {
      long long q = n % p;
      if (q >= k + 1) {
        cout << 1 + n / p << "\n";
        cout << q - k << " ";
        long long cnt = 0;
        for (long long i = q - k + p;; i += p) {
          cout << i << " ";
          cnt++;
          if (cnt == n / p) {
            break;
          }
        }
      } else {
        long long ex = k + 1 - q;
        long long st = k + 1 - ex;
        cout << n / p + 1 << "\n";
        cout << st << " ";
        long long cnt = 0;
        for (long long i = st + p;; i += p) {
          cout << i << " ";
          cnt++;
          if (cnt == n / p) {
            break;
          }
        }
      }
    }
  }
}