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.util.*;
import java.io.*;
public class C{
	public static void main(String[] args)
	{
		FastScanner fs = new FastScanner();
		int n = fs.nextInt(); int k = fs.nextInt();
		List<Integer> temp = new ArrayList();
		boolean[] visited = new boolean[n+1];
		visited[0] = true;
		int start = n-1;
		for(int i=1;i<k;i++)
		{
			temp.add(start);
			start--;
		}
//		for(int i:temp)System.out.print(i+" ");
//		System.out.println();
		List<Integer> ans = new ArrayList();
		ans.add(1);
		int count = 1;
		for(int i:temp)
		{
			int a = ans.get(ans.size()-1);
			if((a+i<=n) && visited[(a+i)%(n+1)]==false) {ans.add((a+i)%(n+1));visited[(a+i)%(n+1)]=true;}
			else {ans.add((a-i)%(n+1));visited[(a-i)%(n+1)]=true;}
			count++;
		}
		while(count<n)
		{
			int a = ans.get(ans.size()-1);
			if(visited[(a+1)%(n+1)]==false) {ans.add((a+1)%(n+1));visited[(a+1)%(n+1)]=true;}
			else {ans.add((a-1)%(n+1));visited[(a-1)%(n+1)]=true;}
			count++;
		}
		for(int i:ans)System.out.print(i+" ");
		System.out.println();
	}


	
	
	
	
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}

	
}




