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. 
import java.io.*;
import java.util.*;

public class A {

	BufferedReader br;
	PrintWriter out;
	StringTokenizer st;
	boolean eof;

	void solve() throws IOException {
		int n = nextInt();
		int k = nextInt();
		int l = 1, r = n;
		int sign = 1;
		for (int i = 0; i < k - 1; i++) {
			if ((i & 1) == 0) {
				out.print((l++) + " ");
			} else {
				out.print((r--) + " ");
			}
			sign = -sign;
		}
		if (sign == 1) {
			for (int i = l; i <= r; i++) {
				out.print(i + " ");
			}
		} else {
			for (int i = r; i >= l; i--) {
				out.print(i + " ");
			}
		}
		out.println();
	}

	A() throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		solve();
		out.close();
	}

	public static void main(String[] args) throws IOException {
		new A();
	}

	String nextToken() {
		while (st == null || !st.hasMoreTokens()) {
			try {
				st = new StringTokenizer(br.readLine());
			} catch (Exception e) {
				eof = true;
				return null;
			}
		}
		return st.nextToken();
	}

	String nextString() {
		try {
			return br.readLine();
		} catch (IOException e) {
			eof = true;
			return null;
		}
	}

	int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}
}