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;
const long long MOD = 1e9 + 7;
long long gcd(long long a, long long b) { return (b == 0 ? a : gcd(b, a % b)); }
long long power(long long x, long long y) {
  long long ans = 1;
  while (y) {
    if (y & 1) ans = (ans * x) % MOD;
    y >>= 1;
    x = (x * x) % MOD;
  }
  return ans % MOD;
}
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  ;
  long long n, k;
  cin >> n >> k;
  long long a = n, b = n - k;
  for (long long i = (long long)0; i < (long long)k / 2; ++i) {
    cout << a << " ";
    --a;
    cout << b << " ";
    ++b;
  }
  if (!(k & 1)) {
    cout << a << " ";
  } else {
    cout << a << " " << b << " ";
  }
  for (long long i = (long long)n - k - 1; i > 0; --i) {
    cout << i << " ";
  }
  cout << '\n';
  return 0;
}
