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;
template <class T>
inline void si(T &x) {
  register int c = getchar();
  x = 0;
  int neg = 0;
  for (; ((c<48 | c> 57) && c != '-'); c = getchar())
    ;
  if (c == '-') {
    neg = 1;
    c = getchar();
  }
  for (; c > 47 && c < 58; c = getchar()) {
    x = (x << 1) + (x << 3) + c - 48;
  }
  if (neg) x = -x;
}
template <class T>
inline T bigmod(T p, T e, T M) {
  long long ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % M;
    p = (p * p) % M;
  }
  return (T)ret;
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T modinverse(T a, T M) {
  return bigmod(a, M - 2, M);
}
void io() {
  freopen("in.txt", "r", stdin);
  freopen("out.txt", "w", stdout);
}
const int N = 1e5 + 5;
set<int> S, pq;
int A[N];
bool vis[N];
vector<int> pp;
int main() {
  int n, k;
  si(n);
  si(k);
  int bb = n;
  for (int i = 0; i < k; i++) S.insert(i + 1);
  A[0] = n;
  vis[n] = 1;
  int foo = -1;
  int i;
  for (i = 1; i < n && S.size(); i++) {
    int oo = *S.rbegin();
    int p = A[i - 1] + foo * oo;
    A[i] = p;
    vis[p] = 1;
    S.erase(*S.rbegin());
    foo *= -1;
  }
  set<int> SS;
  for (int oo = 0; oo < n; oo++)
    if (vis[oo + 1] == 0) {
      SS.insert(oo + 1);
    }
  int kolp = 0;
  for (; i < n; i++) {
    auto d = SS.lower_bound(A[i - 1] - k);
    int oo = *d;
    SS.erase(oo);
    A[i] = oo;
  }
  for (int i = 0; i < n; i++) printf("%d ", A[i]);
  return 0;
}
