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;
void bug() { cout << "here\n"; }
template <typename T>
inline T S(T a) {
  return a * a;
}
template <class T>
inline string tostring(T a) {
  ostringstream os("");
  os << a;
  return os.str();
}
template <typename T>
inline long long tolong(T a) {
  long long res;
  istringstream os(a);
  os >> res;
  return res;
}
template <typename T>
inline T gcd(T a, T b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
template <typename T>
inline T bigmod(T a, T b, T m) {
  if (b == 0)
    return 1;
  else if (b % 2 == 0)
    return S(bigmod(a, b / 2, m)) % m;
  else
    return (a % m * bigmod(a, b - 1, m)) % m;
}
const int inf = (int)1e9 + 5;
const long long linf = (long long)1e16 + 5;
const long long modd = (long long)1e9 + 7;
const int mod = 10000007;
void pr(int f) {
  if (f == 1)
    cout << "YES\n";
  else
    cout << "NO\n";
}
bool isprime(long long n) {
  if (n < 2) return 0;
  if (n == 2)
    return 1;
  else {
    for (int i = 2; i * i <= n; i++)
      if (n % i == 0) return 0;
  }
  return 1;
}
long long countSum(long long x) {
  long long res = 0;
  while (x != 0) {
    res += (x % 10);
    x /= 10;
  }
  return res;
}
long long power(long long a, long long b) {
  if (b == 0) return 1;
  if (b == 1)
    return a;
  else {
    long long res = power(a, b / 2);
    if (b % 2)
      return res * a * res;
    else
      return res * res;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  long long n, k;
  cin >> n >> k;
  long long x = 1, y = n;
  for (long long i = 1; i <= n && i <= k; i++) {
    if (i % 2)
      cout << x++ << " ";
    else
      cout << y-- << " ";
  }
  if (k % 2)
    for (long long i = x; i <= y; i++) {
      cout << i << " ";
    }
  else {
    for (long long i = y; i >= x; i--) {
      cout << i << " ";
    }
  }
}
