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>
long long mpow(long long a, long long n, long long mod) {
  long long ret = 1;
  long long b = a;
  while (n) {
    if (n & 1) ret = (ret * b) % mod;
    b = (b * b) % mod;
    n >>= 1;
  }
  return (long long)ret;
}
long long gcd(long long a, long long b) {
  if (a == 0 || b == 0) return 0;
  if (a == b) return a;
  if (a > b) return gcd(a - b, b);
  return gcd(a, b - a);
}
inline void read(int &x) {
  x = 0;
  register char c = getchar();
  for (; c < '0' || c > '9'; c = getchar())
    ;
  for (; c >= '0' && c <= '9'; c = getchar())
    x = (x << 3) + (x << 1) + (c - '0');
}
inline void write(int x) {
  register char buffor[35];
  register int i = 0;
  do {
    buffor[i++] = (x % 10) + '0';
    x /= 10;
  } while (x);
  i--;
  while (i >= 0) putchar(buffor[i--]);
  putchar('\n');
}
inline int in() {
  int x;
  read(x);
  return x;
}
void solve();
int dbg = 1;
using namespace std;
int main() {
  int t = 1;
  for (int i = 0; i < t; i++) {
    solve();
  }
  return 0;
}
void solve() {
  long long n, k, i, j;
  cin >> n >> k;
  j = 2 * k + 1;
  if (n % j == 0)
    cout << n / j << endl;
  else
    cout << n / j + 1 << endl;
  if (n % j != 0) {
    for (i = max(k + 1 - j + (n % j), 1ll); i <= n; i += j) cout << i << " ";
    cout << endl;
  } else {
    for (i = k + 1; i <= n; i += j) cout << i << " ";
    cout << endl;
  }
}