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:

/**
 * Date: 29 Sep, 2019
 * Link: 
 * 
 * @author Prasad Chaudhari
 * @linkedIn: https://www.linkedin.com/in/prasad-chaudhari-841655a6/
 * @git: https://github.com/Prasad-Chaudhari
 */
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.Arrays;

public class newProgram1 {

    public static void main(String[] args) throws IOException {
        // TODO code application logic here
        FastIO in = new FastIO();
        int n = in.ni();
        int k = in.ni();
        int a[] = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = i;
        }
        int p = solve(a, 0, n, k - 1);
        if (p == -1 || p > 0) {
            System.out.println(-1);
            return;
        }
        for (int i = 0; i < n; i++) {
            in.println(a[i] + 1);
        }
        in.bw.flush();
    }

    private static int solve(int[] a, int l, int r, int k) {
        if (k == -1) {
            return -1;
        } else if (k == 0) {
            return 0;
        }
        // System.out.println(l + " " + r + " " + k + ":");
        // for (int i = l; i < r; i++) {
        //     System.out.print(a[i] + " ");
        // }
        // System.out.println();
        if (l + 1 == r) {
            return k;
        } else {
            if (k == 1) {
                return -1;
            }
            k -= 2;
            int mid = (l + r) / 2;
            a[mid] ^= a[mid - 1];
            a[mid - 1] ^= a[mid];
            a[mid] ^= a[mid - 1];
            int ans = solve(a, l, mid, k);
            if (ans == -1) {
                return -1;
            } else {
                return solve(a, mid, r, ans);
            }
        }
    }

    static class Data implements Comparable<Data> {

        int a, b;

        public Data(int a, int b) {
            this.a = a;
            this.b = b;
        }

        @Override
        public int compareTo(Data o) {
            if (a == o.a) {
                return Integer.compare(b, o.b);
            }
            return Integer.compare(a, o.a);
        }

        public static void sort(int a[]) {
            Data d[] = new Data[a.length];
            for (int i = 0; i < a.length; i++) {
                d[i] = new Data(a[i], 0);
            }
            Arrays.sort(d);
            for (int i = 0; i < a.length; i++) {
                a[i] = d[i].a;
            }
        }
    }

    static class FastIO {

        private final BufferedReader br;
        private final BufferedWriter bw;
        private String s[];
        private int index;

        public FastIO() throws IOException {
            br = new BufferedReader(new InputStreamReader(System.in));
            bw = new BufferedWriter(new OutputStreamWriter(System.out, "UTF-8"));
            s = br.readLine().split(" ");
            index = 0;
        }

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

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

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

        public String next() throws IOException {
            return nextToken();
        }

        public String nli() throws IOException {
            try {
                return br.readLine();
            } catch (IOException ex) {

            }
            return null;
        }

        public int[] gia(int n) throws IOException {
            int a[] = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = ni();
            }
            return a;
        }

        public int[] gia(int n, int start, int end) throws IOException {
            validate(n, start, end);
            int a[] = new int[n];
            for (int i = start; i < end; i++) {
                a[i] = ni();
            }
            return a;
        }

        public double[] gda(int n) throws IOException {
            double a[] = new double[n];
            for (int i = 0; i < n; i++) {
                a[i] = nd();
            }
            return a;
        }

        public double[] gda(int n, int start, int end) throws IOException {
            validate(n, start, end);
            double a[] = new double[n];
            for (int i = start; i < end; i++) {
                a[i] = nd();
            }
            return a;
        }

        public long[] gla(int n) throws IOException {
            long a[] = new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = nl();
            }
            return a;
        }

        public long[] gla(int n, int start, int end) throws IOException {
            validate(n, start, end);
            long a[] = new long[n];
            for (int i = start; i < end; i++) {
                a[i] = nl();
            }
            return a;
        }

        public int[][][] gwtree(int n) throws IOException {
            int m = n - 1;
            int adja[][] = new int[n + 1][];
            int weight[][] = new int[n + 1][];
            int from[] = new int[m];
            int to[] = new int[m];
            int count[] = new int[n + 1];
            int cost[] = new int[n + 1];
            for (int i = 0; i < m; i++) {
                from[i] = i + 1;
                to[i] = ni();
                cost[i] = ni();
                count[from[i]]++;
                count[to[i]]++;
            }
            for (int i = 0; i <= n; i++) {
                adja[i] = new int[count[i]];
                weight[i] = new int[count[i]];
            }
            for (int i = 0; i < m; i++) {
                adja[from[i]][count[from[i]] - 1] = to[i];
                adja[to[i]][count[to[i]] - 1] = from[i];
                weight[from[i]][count[from[i]] - 1] = cost[i];
                weight[to[i]][count[to[i]] - 1] = cost[i];
                count[from[i]]--;
                count[to[i]]--;
            }
            return new int[][][] { adja, weight };
        }

        public int[][][] gwg(int n, int m) throws IOException {
            int adja[][] = new int[n + 1][];
            int weight[][] = new int[n + 1][];
            int from[] = new int[m];
            int to[] = new int[m];
            int count[] = new int[n + 1];
            int cost[] = new int[n + 1];
            for (int i = 0; i < m; i++) {
                from[i] = ni();
                to[i] = ni();
                cost[i] = ni();
                count[from[i]]++;
                count[to[i]]++;
            }
            for (int i = 0; i <= n; i++) {
                adja[i] = new int[count[i]];
                weight[i] = new int[count[i]];
            }
            for (int i = 0; i < m; i++) {
                adja[from[i]][count[from[i]] - 1] = to[i];
                adja[to[i]][count[to[i]] - 1] = from[i];
                weight[from[i]][count[from[i]] - 1] = cost[i];
                weight[to[i]][count[to[i]] - 1] = cost[i];
                count[from[i]]--;
                count[to[i]]--;
            }
            return new int[][][] { adja, weight };
        }

        public int[][] gtree(int n) throws IOException {
            int adja[][] = new int[n + 1][];
            int from[] = new int[n - 1];
            int to[] = new int[n - 1];
            int count[] = new int[n + 1];
            for (int i = 0; i < n - 1; i++) {
                from[i] = i + 1;
                to[i] = ni();
                count[from[i]]++;
                count[to[i]]++;
            }
            for (int i = 0; i <= n; i++) {
                adja[i] = new int[count[i]];
            }
            for (int i = 0; i < n - 1; i++) {
                adja[from[i]][--count[from[i]]] = to[i];
                adja[to[i]][--count[to[i]]] = from[i];
            }
            return adja;
        }

        public int[][] gg(int n, int m) throws IOException {
            int adja[][] = new int[n + 1][];
            int from[] = new int[m];
            int to[] = new int[m];
            int count[] = new int[n + 1];
            for (int i = 0; i < m; i++) {
                from[i] = ni();
                to[i] = ni();
                count[from[i]]++;
                count[to[i]]++;
            }
            for (int i = 0; i <= n; i++) {
                adja[i] = new int[count[i]];
            }
            for (int i = 0; i < m; i++) {
                adja[from[i]][--count[from[i]]] = to[i];
                adja[to[i]][--count[to[i]]] = from[i];
            }
            return adja;
        }

        public void print(String s) throws IOException {
            bw.write(s);
        }

        public void println(String s) throws IOException {
            bw.write(s);
            bw.newLine();
        }

        public void print(int s) throws IOException {
            bw.write(s + "");
        }

        public void println(int s) throws IOException {
            bw.write(s + "");
            bw.newLine();
        }

        public void print(long s) throws IOException {
            bw.write(s + "");
        }

        public void println(long s) throws IOException {
            bw.write(s + "");
            bw.newLine();
        }

        public void print(double s) throws IOException {
            bw.write(s + "");
        }

        public void println(double s) throws IOException {
            bw.write(s + "");
            bw.newLine();
        }

        private String nextToken() throws IndexOutOfBoundsException, IOException {
            if (index == s.length) {
                s = br.readLine().split(" ");
                index = 0;
            }
            return s[index++];
        }

        private void validate(int n, int start, int end) {
            if (start < 0 || end >= n) {
                throw new IllegalArgumentException();
            }
        }
    }
}