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.math.BigInteger;
import java.util.*;
 
 
 
public class Main {
	
	static InputReader in = new InputReader(System.in);
	static PrintWriter out = new PrintWriter(System.out);
	
	static int oo = (int)1e9;
//	static long oo = (long)1e15;
//	static int mod = 1_000_000_007;
	static int mod = 998244353;
	
	static int[] dx = {1, 0, -1, 0};
	static int[] dy = {0, -1, 0, 1};
	
	static int M = 5001;
	static double EPS = 1e-13;
	static int num = 1;
	
	
	public static void main(String[] args) throws IOException {

		int n = in.nextInt();
		int k = in.nextInt();
		int[] a = new int[n];
		boolean p = solve(a, 0, n, k);
		if(!p) {
			System.out.println(-1);
		}
		else {
			for(int x : a)
				System.out.print(x + " ");
		}
		
		out.close();
	}
	
	static boolean solve(int[] a, int l, int r, int k) {
		if(k % 2 == 0)
			return false;
		int len = r - l;
		int maxcall = len * 2 - 1;
		if(k > maxcall)
			return false;
		if(k == 1) {
			for(int i = l; i < r; ++i)
				a[i] = num++;
			return true;
		}
		int mid = (l + r) / 2;
		int leftcall = (mid - l) * 2 - 1;
		if(leftcall >= k - 2) {
			solve(a, mid, r, 1);
			solve(a, l, mid, k - 2);
			return true;
		}
		for(int i = mid-1; i >= l; --i) {
			a[i] = num++;
		}
		solve(a, mid, r, k - 1 - leftcall);
		return true;
	}

	static int find(int[] g, int x) {
		return g[x] = g[x] == x ? x : find(g, g[x]);
	}
	
	static void union(int[] g, int[] size, int x, int y) {
		x = find(g, x); y = find(g, y);
		if(x == y)
			return;
		if(size[x] < size[y]) {
			g[x] = y;
			size[y] += size[x];
		}
		else {
			g[y] = x;
			size[x] += size[y];
		}
	}
	
	static class Segment {
		Segment left, right;
		int size, minIdx;
//		int time, lazy;
		
		public Segment(int[] a, int l, int r) {
			super();
			if(l == r) {
				this.minIdx = l;
				this.size = 1;
				return;
			}
			int mid = (l + r) / 2;
			left = new Segment(a, l, mid);
			right = new Segment(a, mid+1, r);
			if(a[left.minIdx] <= a[right.minIdx])
				this.minIdx = left.minIdx;
			else
				this.minIdx = right.minIdx;
			this.size = left.size + right.size;
		}
		
		boolean covered(int ll, int rr, int l, int r) {
			return ll <= l && rr >= r;
		}
		
		boolean noIntersection(int ll, int rr, int l, int r) {
			return ll > r || rr < l;
		}
		
//		void lazyPropagation() {
//			if(lazy != 0) {
//				if(left != null) {
//					left.setLazy(this.time, this.lazy);
//					right.setLazy(this.time, this.lazy);
//				}
//				else {
//					val = lazy;
//				}
//			}
//		}
		
//		void setLazy(int time, int lazy) {
//			if(this.time != 0 && this.time <= time)
//				return;
//			this.time = time;
//			this.lazy = lazy;
//		}
		
		
		int querySize(int ll, int rr, int l, int r) {
//			lazyPropagation();
			if(noIntersection(ll, rr, l, r))
				return 0;
			if(covered(ll, rr, l, r))
				return size;
			int mid = (l + r) / 2;
			int leftSize = left.querySize(ll, rr, l, mid);
			int rightSize = right.querySize(ll, rr, mid+1, r);
			return leftSize + rightSize;
		}
		
		int query(int k, int l, int r) {
			Segment trace = this;
			while(l < r) {
				int mid = (l + r) / 2;
				if(trace.left.size > k) {
					trace = trace.left;
					r = mid;
				}
				else {
					k -= trace.left.size;
					trace = trace.right;
					l = mid + 1;
				}
			}
			return l;
		}
		
		void update(int ll, int rr, int l, int r) {
//			lazyPropagation();
			if(noIntersection(ll, rr, l, r))
				return;
			if(covered(ll, rr, l, r)) {
//				setLazy(time, knight);
				this.minIdx = -1;
				this.size = 0;
				return;
			}
			int mid = (l + r) / 2;
			left.update(ll, rr, l, mid);
			right.update(ll, rr, mid+1, r);
			this.size = left.size + right.size;
		}
	}
	
	static long pow(long a, long n, long mod) {
		if(n == 0)
			return 1;
		if(n % 2 == 1)
			return a * pow(a, n-1, mod) % mod;
		long x = pow(a, n / 2, mod);
		return x * x % mod;
	}
	
	
	static int[] getPi(char[] a) {
		int m = a.length;
		int j = 0;
		int[] pi = new int[m];
		for(int i = 1; i < m; ++i) {
			while(j > 0 && a[i] != a[j])
				j = pi[j-1];
			if(a[i] == a[j]) {
				pi[i] = j + 1;
				j++;
			}
		}
		return pi;
	}
	
	static long lcm(long a, long b) {
		return a * b / gcd(a, b);
	}
	
	static BigInteger lcm2(long a, long b) {
		long g = gcd(a, b);
		BigInteger gg = BigInteger.valueOf(g);
		BigInteger aa = BigInteger.valueOf(a);
		BigInteger bb = BigInteger.valueOf(b);
		return aa.multiply(bb).divide(gg);
	}
	
	static boolean nextPermutation(int[] a) {
		for(int i = a.length - 2; i >= 0; --i) {
			if(a[i] < a[i+1]) {
				for(int j = a.length - 1; ; --j) {
					if(a[i] < a[j]) {
						int t = a[i];
						a[i] = a[j];
						a[j] = t;
						for(i++, j = a.length - 1; i < j; ++i, --j) {
							t = a[i];
							a[i] = a[j];
							a[j] = t;
						}
						return true;
					}
				}
			}
		}
		return false;
	}
	
	static void shuffle(Object[] a) {
		Random r = new Random();
		for(int i = a.length - 1; i > 0; --i) {
			int si = r.nextInt(i);
			Object t = a[si];
			a[si] = a[i];
			a[i] = t;
		}
	}
	
	static void shuffle(int[] a) {
		Random r = new Random();
		for(int i = a.length - 1; i > 0; --i) {
			int si = r.nextInt(i);
			int t = a[si];
			a[si] = a[i];
			a[i] = t;
		}
	}
	
	static void shuffle(long[] a) {
		Random r = new Random();
		for(int i = a.length - 1; i > 0; --i) {
			int si = r.nextInt(i);
			long t = a[si];
			a[si] = a[i];
			a[i] = t;
		}
	}
	
	static int lower_bound(int[] a, int n, int k) {
		int s = 0;
		int e = n;
		int m;
		while (e - s > 0) {
			m = (s + e) / 2;
			if (a[m] < k)
				s = m + 1;
			else
				e = m;
		}
		return e;
	}
	static int lower_bound(long[] a, int n, long k) {
		int s = 0;
		int e = n;
		int m;
		while (e - s > 0) {
			m = (s + e) / 2;
			if (a[m] < k)
				s = m + 1;
			else
				e = m;
		}
		return e;
	}
	
	static int gcd(int a, int b) {
		return b == 0 ? a : gcd(b, a % b);
	}
	static long gcd(long a, long b) {
		return b == 0 ? a : gcd(b, a % b);
	}
	static BigInteger gcd(BigInteger a, BigInteger b) {
		return b.compareTo(BigInteger.ZERO) == 0 ? a : gcd(b, a.mod(b));
	}
	
	static class Pair implements Comparable<Pair> {
		int first, second;
 
		public Pair(int first, int second) {
			super();
			this.first = first;
			this.second = second;
		}
 
		@Override
		public int compareTo(Pair o) {
			return this.first != o.first ? this.first - o.first : this.second - o.second;
		}
		
//		@Override
//		public int compareTo(Pair o) {
//			return this.first != o.first ? o.first - this.first : o.second - this.second;
//		}
 
		@Override
		public int hashCode() {
			final int prime = 31;
			int result = 1;
			result = prime * result + first;
			result = prime * result + second;
			return result;
		}
 
		@Override
		public boolean equals(Object obj) {
			if (this == obj)
				return true;
			if (obj == null)
				return false;
			if (getClass() != obj.getClass())
				return false;
			Pair other = (Pair) obj;
			if (first != other.first)
				return false;
			if (second != other.second)
				return false;
			return true;
		}
	}
}
 
 
 
class InputReader {
 
	private final InputStream stream;
	private final byte[] buf = new byte[8192];
	private int curChar, snumChars;
 
	public InputReader(InputStream st) {
		this.stream = st;
	}
 
	public int read() {
		if (snumChars == -1)
			throw new InputMismatchException();
		if (curChar >= snumChars) {
			curChar = 0;
			try {
				snumChars = stream.read(buf);
			} catch (IOException e) {
				throw new InputMismatchException();
			}
			if (snumChars <= 0)
				return -1;
		}
		return buf[curChar++];
	}
 
	public int nextInt() {
		int c = read();
		while (isSpaceChar(c)) {
			c = read();
		}
		int sgn = 1;
		if (c == '-') {
			sgn = -1;
			c = read();
		}
		int res = 0;
		do {
			res *= 10;
			res += c - '0';
			c = read();
		} while (!isSpaceChar(c));
		return res * sgn;
	}
 
	public long nextLong() {
		int c = read();
		while (isSpaceChar(c)) {
			c = read();
		}
		int sgn = 1;
		if (c == '-') {
			sgn = -1;
			c = read();
		}
		long res = 0;
		do {
			res *= 10;
			res += c - '0';
			c = read();
		} while (!isSpaceChar(c));
		return res * sgn;
	}
 
	public int[] nextIntArray(int n) {
		int a[] = new int[n];
		for (int i = 0; i < n; i++) {
			a[i] = nextInt();
		}
		return a;
	}
 
	public String readString() {
		int c = read();
		while (isSpaceChar(c)) {
			c = read();
		}
		StringBuilder res = new StringBuilder();
		do {
			res.appendCodePoint(c);
			c = read();
		} while (!isSpaceChar(c));
		return res.toString();
	}
 
	public String nextLine() {
		int c = read();
		while (isSpaceChar(c))
			c = read();
		StringBuilder res = new StringBuilder();
		do {
			res.appendCodePoint(c);
			c = read();
		} while (!isEndOfLine(c));
		return res.toString();
	}
 
	public boolean isSpaceChar(int c) {
		return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
	}
 
	private boolean isEndOfLine(int c) {
		return c == '\n' || c == '\r' || c == -1;
	}
 
}