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.util.StringTokenizer;

public class Main {

    static MyScanner in;
    static PrintWriter out;

    static int cnt = 1, k;

    public static void main(String[] args) throws IOException {
        in = new MyScanner();
//        out = new PrintWriter(new BufferedOutputStream(System.out), false);

        int n = in.nextInt();
        k = in.nextInt();
        if (k % 2 == 0) {
            System.out.println(-1);
            return;
        }

        int[] arr = new int[n];
        for (int i = 0; i < n; i++)
            arr[i] = i + 1;

        merge(arr, 0, n);

        if (cnt != k) {
            System.out.println(-1);
        } else {
            for (int a : arr) {
                System.out.print(a);
                System.out.print(" ");
            }
            System.out.println();
        }
    }

    static void merge(int[] arr, int lo, int hi) {
        if (lo >= hi - 1)
            return;
        if (cnt == k)
            return;

        int mid = (lo + hi) / 2;
        swap(arr, mid, mid - 1);
        cnt += 2;

        if (cnt != k)
            merge(arr, lo, mid);
        if (cnt != k)
            merge(arr, mid, hi);
    }

    static void swap(int[] arr, int i, int j) {
        int t = arr[i];
        arr[i] = arr[j];
        arr[j] = t;
    }

}

class MyScanner implements Closeable {

    public static final String CP1251 = "cp1251";

    private final BufferedReader br;
    private StringTokenizer st;
    private String last;

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

    public MyScanner(String path) throws IOException {
        br = new BufferedReader(new FileReader(path));
    }

    public MyScanner(String path, String decoder) throws IOException {
        br = new BufferedReader(new InputStreamReader(new FileInputStream(path), decoder));
    }

    public void close() throws IOException {
        br.close();
    }

    public String next() throws IOException {
        while (st == null || !st.hasMoreElements())
            st = new StringTokenizer(br.readLine());
        last = null;
        return st.nextToken();
    }

    public String next(String delim) throws IOException {
        while (st == null || !st.hasMoreElements())
            st = new StringTokenizer(br.readLine());
        last = null;
        return st.nextToken(delim);
    }

    public String nextLine() throws IOException {
        st = null;
        return (last == null) ? br.readLine() : last;
    }

    public boolean hasNext() {
        if (st != null && st.hasMoreElements())
            return true;

        try {
            while (st == null || !st.hasMoreElements()) {
                last = br.readLine();
                st = new StringTokenizer(last);
            }
        } catch (Exception e) {
            return false;
        }

        return true;
    }

    public String[] nextStrings(int n) throws IOException {
        String[] arr = new String[n];
        for (int i = 0; i < n; i++)
            arr[i] = next();
        return arr;
    }

    public String[] nextLines(int n) throws IOException {
        String[] arr = new String[n];
        for (int i = 0; i < n; i++)
            arr[i] = nextLine();
        return arr;
    }

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

    public int[] nextInts(int n) throws IOException {
        int[] arr = new int[n];
        for (int i = 0; i < n; i++)
            arr[i] = nextInt();
        return arr;
    }

    public Integer[] nextIntegers(int n) throws IOException {
        Integer[] arr = new Integer[n];
        for (int i = 0; i < n; i++)
            arr[i] = nextInt();
        return arr;
    }

    public int[][] next2Ints(int n, int m) throws IOException {
        int[][] arr = new int[n][m];
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++)
                arr[i][j] = nextInt();
        return arr;
    }

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

    public long[] nextLongs(int n) throws IOException {
        long[] arr = new long[n];
        for (int i = 0; i < n; i++)
            arr[i] = nextLong();
        return arr;
    }

    public long[][] next2Longs(int n, int m) throws IOException {
        long[][] arr = new long[n][m];
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++)
                arr[i][j] = nextLong();
        return arr;
    }

    public double nextDouble() throws IOException {
        return Double.parseDouble(next().replace(',', '.'));
    }

    public double[] nextDoubles(int size) throws IOException {
        double[] arr = new double[size];
        for (int i = 0; i < size; i++)
            arr[i] = nextDouble();
        return arr;
    }

    public double[][] next2Doubles(int n, int m) throws IOException {
        double[][] arr = new double[n][m];
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++)
                arr[i][j] = nextDouble();
        return arr;
    }

    public boolean nextBool() throws IOException {
        String s = next();
        if (s.equalsIgnoreCase("true") || s.equals("1"))
            return true;

        if (s.equalsIgnoreCase("false") || s.equals("0"))
            return false;

        throw new IOException("Boolean expected, String found!");
    }
}