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>
double const EPS = 3e-8;
using namespace std;
template <class T>
T gcd(T a, T b) {
  return (b != 0 ? gcd<T>(b, a % b) : a);
}
template <class T>
T lcm(T a, T b) {
  return (a / gcd<T>(a, b) * b);
}
template <class T>
inline bool read(T &x) {
  int c = getchar();
  int sgn = 1;
  while (~c && c < '0' || c > '9') {
    if (c == '-') sgn = -1;
    c = getchar();
  }
  for (x = 0; ~c && '0' <= c && c <= '9'; c = getchar()) x = x * 10 + c - '0';
  x *= sgn;
  return ~c;
}
long long n, k;
const int MX = 1e6 + 9;
long long Ans[MX];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  if (k == 1) {
    for (int i = 0; i < n; i++) {
      if (i) cout << " ";
      cout << i + 1;
    }
    cout << endl;
    return 0;
  }
  long long low = 1, high = k + 1;
  int idx = k, i, j;
  while (1) {
    Ans[idx] = high;
    high--;
    idx--;
    if (idx < 0) break;
    Ans[idx] = low;
    low++;
    idx--;
    if (idx < 0) break;
  }
  for (i = k + 1, j = k + 2; i < n; i++, j++) {
    Ans[i] = j;
  }
  for (i = 0; i < n; i++) {
    if (i > 0) cout << " ";
    cout << Ans[i];
  }
  cout << endl;
  return 0;
}
