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;
inline void inp(int &n) {
  n = 0;
  int ch = getchar();
  int sign = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') sign = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    n = (n << 3) + (n << 1) + ch - '0', ch = getchar();
  n = n * sign;
}
long long gcd(long long A, long long B) {
  return A > B ? gcd(B, A) : A ? gcd(B % A, A) : B;
}
long long lcm(long long A, long long B) { return A / gcd(A, B) * B; }
long long pow(long long A, long long B, long long P) {
  if (!B) return 1 % P;
  long long h = pow(A, B / 2, P);
  h *= h;
  h %= P;
  if (B % 2) h *= A;
  return h % P;
}
vector<int> z_function(string s) {
  int n = (int)s.length();
  vector<int> z(n);
  for (int i = 1, l = 0, r = 0; i < n; ++i) {
    if (i <= r) z[i] = min(r - i + 1, z[i - l]);
    while (i + z[i] < n && s[z[i]] == s[i + z[i]]) ++z[i];
    if (i + z[i] - 1 > r) l = i, r = i + z[i] - 1;
  }
  return z;
}
int tst;
void get_ready() { tst = 1; }
bool f[262145];
vector<int> g[262145], c[262145];
vector<int> ans1, ans2, ans3;
string s, q, s1, s2;
pair<int, int> P[262145];
int a[262145], b[262145];
char D[2555][2555];
int n, m, i, j, k, l, r, x, y, u, v;
long long ans, res, cur;
bool READ() {
  cin >> n >> k;
  for (i = 1; i <= n - k; i++) cout << i << " ";
  r = n;
  l = n - k + 1;
  i = 0;
  for (;;) {
    i++;
    if (l > r) break;
    if (i % 2)
      cout << r << " ", r--;
    else
      cout << l << " ", l++;
  }
  cout << endl;
  return 1;
}
bool SOLVE() { return 1; }
bool PRINT() { return 1; }
bool CLEAR() { return 1; }
int main() {
  get_ready();
  while (tst--) {
    if (!READ()) {
      return 0;
    }
    if (!SOLVE()) {
      return 0;
    }
    if (!PRINT()) {
      return 0;
    }
    if (!CLEAR()) {
      return 0;
    }
  }
}
