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;
vector<int> V;
int main() {
  int N, k, i, Ans;
  scanf("%d%d", &N, &k);
  if (k == 0) {
    printf("%d\n", N);
    for (i = 1; i <= N; ++i) printf("%d ", i);
    puts("");
    return 0;
  }
  if (N <= 2 * k + 1) {
    puts("1");
    if (N <= k + 1) {
      puts("1");
      return 0;
    }
    printf("%d\n", k + 1);
    return 0;
  }
  int a = 2 * k + 1;
  int b = N % a;
  if (b > k || b == 0) {
    if (b == 0)
      Ans = N / a;
    else
      Ans = (N / a) + 1;
    printf("%d\n", Ans);
    for (i = k + 1; i <= N; i += 2 * k + 1) printf("%d ", i);
    puts("");
    return 0;
  }
  int y = b;
  int p = 1;
  int q = y;
  V.emplace_back(1);
  int high = N - k - q;
  for (i = p + 2 * k + 1; i <= high; i += 2 * k + 1) V.emplace_back(i);
  V.emplace_back(N - q + 1);
  printf("%d\n", int(V.size()));
  for (i = 0; i < V.size(); ++i) printf("%d ", V[i]);
  puts("");
  return 0;
}