Permutation p is an ordered set of integers p1, p2, ..., pn, consisting of n distinct positive integers not larger than n. We'll denote as n the length of permutation p1, p2, ..., pn.

Your task is to find such permutation p of length n, that the group of numbers |p1 - p2|, |p2 - p3|, ..., |pn - 1 - pn| has exactly k distinct elements.

Input

The single line of the input contains two space-separated positive integers n, k (1 ≤ k < n ≤ 105).

Output

Print n integers forming the permutation. If there are multiple answers, print any of them.

Examples

Input

3 2


Output

1 3 2


Input

3 1


Output

1 2 3


Input

5 2


Output

1 3 2 4 5

Note

By |x| we denote the absolute value of number x. 
#include <bits/stdc++.h>
using namespace std;
set<int> s;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  int k, n;
  cin >> k >> n;
  for (long long(i) = (2); (i) < (k + 1); (i)++) s.insert(i);
  vector<int> v;
  v.push_back(1);
  for (long long(i) = (n); (i) >= (1); (i)--) {
    if (((int)v.size()) == k) break;
    int id = ((int)v.size()) - 1;
    int a = v[id] - i, b = v[id] + i;
    if (a <= 0 || a > k) a = -1;
    if (b <= 0 || b > k) b = -1;
    int fa = 0, fb = 0;
    if (s.find(a) != s.end()) fa++;
    if (s.find(b) != s.end()) fb++;
    if (fa && fb) {
      int u = min(a, b);
      s.erase(u);
      v.push_back(u);
    } else if (fa && !fb) {
      s.erase(a);
      v.push_back(a);
    } else if (!fa && fb) {
      s.erase(b);
      v.push_back(b);
    }
  }
  while (((int)v.size()) < k) {
    int id = ((int)v.size()) - 1;
    set<int>::iterator it = s.lower_bound(v[id]);
    int u = *it;
    v.push_back(u);
    s.erase(it);
  }
  for (long long(i) = (0); (i) < (((int)v.size())); (i)++) {
    cout << v[i];
    if (i != ((int)v.size()) - 1)
      cout << " ";
    else
      cout << endl;
  }
  return 0;
}
