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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Collection;
import java.util.Collections;
import java.util.List;
import java.util.Scanner;
import java.util.StringTokenizer;
import java.util.*;

public class Q1 {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        ArrayList<Integer> readIntArrayList(int n) {
            ArrayList<Integer> a = new ArrayList<>();
            for (int i = 0; i < n; i++)
                a.add(nextInt());
            return a;
        }

        ArrayList<Long> readLongArray(int n) {
            ArrayList<Long> a = new ArrayList<>();
            for (int i = 0; i < n; i++)
                a.add(nextLong());
            return a;
        }

        int[] readIntArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    static Boolean[] sieveOfEratosthenes(int n, Boolean[] prime) {
        for (int i = 0; i <= n; i++)
            prime[i] = true;

        for (int p = 2; p * p <= n; p++) {
            // If prime[p] is not changed, then it is a
            // prime
            if (prime[p] == true) {
                // Update all multiples of p
                for (int i = p * p; i <= n; i += p)
                    prime[i] = false;
            }
        }
        return prime;

        // Print all prime numbers
    }

    static int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }

    static void p(String s) {
        System.out.print(s);
    }

    static void pln(String s) {
        System.out.println(s);
    }

    public static void main(String[] args) {

        FastReader fr = new FastReader();
        int n = fr.nextInt(), k = fr.nextInt();
        HashSet<Integer> numbers = new HashSet<>();
        int i = 1;
        int sign = 1;
        System.out.print(1 + " ");
        while (k != 1) {
            i = Math.abs(i + sign * k);
            numbers.add(i);
            System.out.print(i + " ");
            sign *= -1;
            k--;
        }
        for (i = 2; i <= n; i++) {
            if (!numbers.contains(i))
                System.out.print(i + " ");
        }

    }

}
