Problem: Merge sort is a well-known sorting algorithm. The main function that sorts the elements of array a with indices from [l, r) can be implemented as follows:

  1. If the segment [l, r) is already sorted in non-descending order (that is, for any i such that l ≤ i < r - 1 a[i] ≤ a[i + 1]), then end the function call; 
  2. Let <image>; 
  3. Call mergesort(a, l, mid); 
  4. Call mergesort(a, mid, r); 
  5. Merge segments [l, mid) and [mid, r), making the segment [l, r) sorted in non-descending order. The merge algorithm doesn't call any other functions. 



The array in this problem is 0-indexed, so to sort the whole array, you need to call mergesort(a, 0, n).

The number of calls of function mergesort is very important, so Ivan has decided to calculate it while sorting the array. For example, if a = {1, 2, 3, 4}, then there will be 1 call of mergesort — mergesort(0, 4), which will check that the array is sorted and then end. If a = {2, 1, 3}, then the number of calls is 3: first of all, you call mergesort(0, 3), which then sets mid = 1 and calls mergesort(0, 1) and mergesort(1, 3), which do not perform any recursive calls because segments (0, 1) and (1, 3) are sorted.

Ivan has implemented the program that counts the number of mergesort calls, but now he needs to test it. To do this, he needs to find an array a such that a is a permutation of size n (that is, the number of elements in a is n, and every integer number from [1, n] can be found in this array), and the number of mergesort calls when sorting the array is exactly k.

Help Ivan to find an array he wants!

Input

The first line contains two numbers n and k (1 ≤ n ≤ 100000, 1 ≤ k ≤ 200000) — the size of a desired permutation and the number of mergesort calls required to sort it.

Output

If a permutation of size n such that there will be exactly k calls of mergesort while sorting it doesn't exist, output  - 1. Otherwise output n integer numbers a[0], a[1], ..., a[n - 1] — the elements of a permutation that would meet the required conditions. If there are multiple answers, print any of them.

Examples

Input

3 3


Output

2 1 3 

Input

4 1


Output

1 2 3 4 

Input

5 6


Output

-1

Solution (in JAVA):

Solution:
import java.io.*;
import java.time.Duration;
import java.time.Instant;
import java.util.*;

/**
 * @author Vishal Singh
 */


public class Main {
    public static void main(String[] args) throws IOException {
        Solver newProblem = new Solver();
        newProblem.run();
    }
}

class Solver extends Vishal {
    static final boolean MULTIPLE_TEST_CASE = false;

    static class Count {
        int callsMade = 1;
    }

    public void solve(int testCaseNumber) throws IOException {
        /* Write your code here LMAO */
        int n = nextInt();
        int k = nextInt();
        if (((k & 1) == 0) || (k > n*2-1)) print(-1);
        else {
            int[] arr = new int[n];
            for (int i = 0; i < n; i++) {
                arr[i] = i + 1;
            }
            Count c = new Count();
            unsort(arr, 0, n, k, c);
            StringBuilder ans = new StringBuilder();
            for (int i = 0; i < n; i++) {
                ans.append(arr[i]).append(" ");
            }
            print(ans);
        }
        /* _________________________ */
    }

    public void unsort(int[] arr, int low, int high, int k, Count c) {
        if (c.callsMade == k) return;

        if (high - low == 1) return;

        int mid = low + high >>> 1;

        int temp = arr[mid];
        arr[mid] = arr[mid - 1];
        arr[mid - 1] = temp;

        c.callsMade += 2;

        unsort(arr, low, mid, k, c);
        unsort(arr, mid , high, k, c);
    }

    public void run() throws IOException {
        int t = MULTIPLE_TEST_CASE ? nextInt() : 1;
        Instant start = Instant.now();
        for (int i = 0; i < t; i++) {
            try {
                solve(i + 1);
            } catch (Exception e) {
                e.printStackTrace();
            }
        }
        Instant end = Instant.now();
        Duration timeElapsed = Duration.between(start, end);
        //println(timeElapsed.toMillis()+" ms");
    }
}

class Vishal extends FastIO {
    public static final int MOD = 1000000007;
    public static final int MAX = Integer.MAX_VALUE;
    public static final int MIN = Integer.MIN_VALUE;

    public static boolean isPrime(long number) {
        if (number <= 1) return false;
        if (number == 2 || number == 3) return true;
        if (number % 2 == 0 || number % 3 == 0) return false;
        for (int i = 5; i * i <= number; i += 6) {
            if (number % i == 0 || number % (i + 2) == 0) return false;
        }
        return true;
    }

    /*
     * Time Complexity: O(Log min(a, b))
     * */
    public static long hcf(long a, long b) {
        while (true) {
            if (a == 0) {
                return b;
            }
            long a1 = a;
            a = b % a1;
            b = a1;
        }
    }

    public static long lcm(long a, long b) {
        return (a / hcf(a, b)) * b;
    }

    /**
     * @param arr The array whose value is to be swapped.
     * @param i   The first index to swap with.
     * @param j   The second index to swap with.
     */
    public static void swap(int[] arr, int i, int j) {
        int temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
    }

    public static class Pair<A, B> {
        A first;
        B second;

        Pair(A first, B second) {
            this.first = first;
            this.second = second;
        }

        @Override
        public String toString() {
            return "Pair{" +
                    "first=" + first +
                    ", second=" + second +
                    '}';
        }
    }

    public static int countDigit(long n) {
        return (int) Math.floor(Math.log10(n) + 1);
    }

}

abstract class FastIO implements Output, Input {
    static final BufferedReader inputReader = new BufferedReader(new InputStreamReader(System.in));
    static final BufferedWriter outputWriter = new BufferedWriter(new OutputStreamWriter(System.out));
    static StringTokenizer st;

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

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

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

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

    final public String nextLine() throws IOException {
        return inputReader.readLine();
    }

    final public void println() throws IOException {
        outputWriter.write("\n");
        outputWriter.flush();
    }

    final public void println(Object o) throws IOException {
        outputWriter.write(o + "\n");
        outputWriter.flush();
    }

    final public void print(Object o) throws IOException {
        outputWriter.write(String.valueOf(o));
        outputWriter.flush();
    }
}

interface Input {
    BufferedReader inputReader = null;
    BufferedWriter outputWriter = null;
    StringTokenizer st = null;

    String next();

    int nextInt();

    long nextLong();

    double nextDouble();

    String nextLine() throws IOException;
}

interface Output {
    void println() throws IOException;

    void println(Object o) throws IOException;

    void print(Object o) throws IOException;
}