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.util.*;
import java.text.*;
import java.io.*;
import java.math.*;
public class code5 {
	InputStream is;
	PrintWriter out;
	static long mod=pow(10,9)+7;
	static int dx[]={0,0,1,-1},dy[]={1,-1,0,0};
	long req[];
 	void solve() throws Exception
	{
 		int n=ni();
 		int k=ni();
 		if(k%2==0||k>count(0,n-1))
 			out.print(-1);
 		else 
 		{
 			int arr[]=new int[n+1];
 			for(int i=0;i<n;i++)
 				arr[i]=i;
 			this.k=k-1;
 			ans(arr,0,n);
 			for(int i=0;i<arr.length-1;i++)
 				out.print((arr[i]+1)+" ");
 		}
 			
	}
 	int k;
 	void ans(int arr[],int l,int r)
 	{
 		if(k==0)
 			return;
 		if(l+1==r)
 			return;
 		int mid=(l+r)/2;
 		int temp=arr[mid-1];
 		arr[mid-1]=arr[mid];
 		arr[mid]=temp;
 		k-=2;
 		ans(arr,l,mid);
 		ans(arr,mid,r);
 	}
 	boolean check(int arr[],int l,int r)
 	{
 		for(int i=l+1;i<=r;i++)
 		{
 			if(arr[i]<arr[i-1])
 				return false;
 		}
 		return true;
 	}
 	int count(int l,int r)
 	{
 		if(l==r)
 			return 1;
 		if(l>r)
 			return 0;
 		int mid=(l+r)/2;
 		return 1+count(l,mid)+count(mid+1,r);
 	}
 	ArrayList<Pair> al[];
 	void take(int n,int m)
 	{
 		al=new ArrayList[n];
 		for(int i=0;i<n;i++)
 			al[i]=new ArrayList<Pair>();
 		for(int i=0;i<m;i++)
 		{
 			int x=ni()-1;
 			int y=ni()-1;
 			int w=ni();
 			al[x].add(new Pair(y,w,-1));
 			al[y].add(new Pair(x,w,-1));
 		}
 	}
 	int check(long n)
 	{
 		int count=0;
 		while(n!=0)
 		{
 			if(n%10!=0)
 				break;
 			n/=10;
 			count++;
 		}
 		return count;
 	}
	public static int count(int x)
	{
		int num=0;
		while(x!=0)
		{
			x=x&(x-1);
			num++;
		}
		return num;
	}
	static long d, x, y;
	void extendedEuclid(long A, long B) {
	    if(B == 0) {
	        d = A;
	        x = 1;
	        y = 0;
	    }
	    else {
	        extendedEuclid(B, A%B);
	        long temp = x;
	        x = y;
	        y = temp - (A/B)*y;
	    }
	}
	long modInverse(long A,long M) //A and M are coprime
	{
	    extendedEuclid(A,M);
	    return (x%M+M)%M;    //x may be negative
	}
	public static void mergeSort(int[] arr, int l ,int r){
		if((r-l)>=1){
			int mid = (l+r)/2;
			mergeSort(arr,l,mid);
			mergeSort(arr,mid+1,r);
			merge(arr,l,r,mid);
		}
	}
	public static void merge(int arr[], int l, int r, int mid){
		int n1 = (mid-l+1), n2 = (r-mid);
		int left[] = new int[n1];
		int right[] = new int[n2];
		for(int i =0 ;i<n1;i++) left[i] = arr[l+i];
		for(int i =0 ;i<n2;i++) right[i] = arr[mid+1+i];
		int i =0, j =0, k = l;
		while(i<n1 && j<n2){
			if(left[i]>right[j]){
				arr[k++] = right[j++];
			}
			else{
				arr[k++] = left[i++];
			}
		}
		while(i<n1) arr[k++] = left[i++];
		while(j<n2) arr[k++] = right[j++];
	}
	public static void mergeSort(long[] arr, int l ,int r){
		if((r-l)>=1){
			int mid = (l+r)/2;
			mergeSort(arr,l,mid);
			mergeSort(arr,mid+1,r);
			merge(arr,l,r,mid);
		}
	}
	public static void merge(long arr[], int l, int r, int mid){
		int n1 = (mid-l+1), n2 = (r-mid);
		long left[] = new long[n1];
		long right[] = new long[n2];
		for(int i =0 ;i<n1;i++) left[i] = arr[l+i];
		for(int i =0 ;i<n2;i++) right[i] = arr[mid+1+i];
		int i =0, j =0, k = l;
		while(i<n1 && j<n2){
			if(left[i]>right[j]){
				arr[k++] = right[j++];
			}
			else{
				arr[k++] = left[i++];
			}
		}
		while(i<n1) arr[k++] = left[i++];
		while(j<n2) arr[k++] = right[j++];
	}
	 static class Pair implements Comparable<Pair>{
		 
	       int x,y,k;
	       int i,dir;
		Pair (int xx,int yy,int k){
			this.x=xx;
			this.y=yy;
			this.k=k;
		}   
		public int compareTo(Pair o) {
			if(this.y!=o.y)
				return this.y-o.y;
			return this.k-o.k;
		}
	 
		public String toString(){
			return x+" "+y+" "+k+" "+i;}
		public boolean equals(Object o) {
			if (o instanceof Pair) {
				Pair p = (Pair)o;
				return p.x == x && p.y == y;
			}
			return false;
		}
		public int hashCode() {
			return new Long(x).hashCode()*31 + new Long(y).hashCode() ;
		}
	} 
	    
	    public static boolean isPal(String s){
	        for(int i=0, j=s.length()-1;i<=j;i++,j--){
	                if(s.charAt(i)!=s.charAt(j)) return false;
	        }
	        return true;
	    }
	    public static String rev(String s){
			StringBuilder sb=new StringBuilder(s);
			sb.reverse();
			return sb.toString();
	    }
	    
	    public static long gcd(long x,long y){
		if(y==0)
			return x;
		else
			return gcd(y,x%y);
	    }
	    
	    public static int gcd(int x,int y){
	    	if(y==0) 
	    		return x; 
	    	return gcd(y,x%y);
	    }
	    
	    public static long gcdExtended(long a,long b,long[] x){
	        
	        if(a==0){
	            x[0]=0;
	            x[1]=1;
	            return b;
	        }
	        long[] y=new long[2];
	        long gcd=gcdExtended(b%a, a, y);
	        
	        x[0]=y[1]-(b/a)*y[0];
	        x[1]=y[0];
	        
	        return gcd;
	    }
	    
	    public static int abs(int a,int b){
		return (int)Math.abs(a-b);
	    }
	 
	    public static long abs(long a,long b){
		return (long)Math.abs(a-b);
	    }
	    
	    public static int max(int a,int b){
		if(a>b)
			return a;
		else
			return b;
	    }
	 
	    public static int min(int a,int b){
		if(a>b)
			return b;
		else 
			return a;
	    }
	    
	    public static long max(long a,long b){
		if(a>b)
			return a;
		else
			return b;
	    }
	 
	    public static long min(long a,long b){
		if(a>b)
			return b;
		else 
			return a;
	    }
	 
	    public static long pow(long n,long p,long m){
		 long  result = 1;
		  if(p==0)
		    return 1;
		if (p==1)
		    return n;
		while(p!=0)
		{
		    if(p%2==1)
		        result *= n;
		    if(result>=m)
		    result%=m;
		    p >>=1;
		    n*=n;
		    if(n>=m)
		    n%=m;
		}
		return result;
	    }
	    
	    public static long pow(long n,long p){
		long  result = 1;
		  if(p==0)
		    return 1;
		if (p==1)
		    return n;
		while(p!=0)
		{
		    if(p%2==1)
		        result *= n;	    
		    p >>=1;
		    n*=n;	    
		}
		return result;
	    }
	    public static void debug(Object... o) {
			System.out.println(Arrays.deepToString(o));
		}
	    void run() throws Exception {
			is = System.in;
			out = new PrintWriter(System.out);
			solve();
			out.flush();
		}
	   
	    public static void main(String[] args) throws Exception {
			new Thread(null, new Runnable() {
				public void run() {
					try {
						new code5().run();
					} catch (Exception e) {
						e.printStackTrace();
					}
				}
			}, "1", 1 << 26).start();
			
	    	//new code5().run();
		}
	    private byte[] inbuf = new byte[1024];
		public int lenbuf = 0, ptrbuf = 0;
	 
		private int readByte() {
			if (lenbuf == -1)
				throw new InputMismatchException();
			if (ptrbuf >= lenbuf) {
				ptrbuf = 0;
				try {
					lenbuf = is.read(inbuf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (lenbuf <= 0)
					return -1;
			}
			return inbuf[ptrbuf++];
		}
	 
		private boolean isSpaceChar(int c) {
			return !(c >= 33 && c <= 126);
		}
	 
		private int skip() {
			int b;
			while ((b = readByte()) != -1 && isSpaceChar(b));
			return b;
		}
	 
		private double nd() {
			return Double.parseDouble(ns());
		}
	 
		private char nc() {
			return (char) skip();
		}
	 
		private String ns() {
			int b = skip();
			StringBuilder sb = new StringBuilder();
			while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != ' ')
				sb.appendCodePoint(b);
				b = readByte();
			}
			return sb.toString();
		}
	 
		private char[] ns(int n) {
			char[] buf = new char[n];
			int b = skip(), p = 0;
			while (p < n && !(isSpaceChar(b))) {
				buf[p++] = (char) b;
				b = readByte();
			}
			return n == p ? buf : Arrays.copyOf(buf, p);
		}
	 
		private char[][] nm(int n, int m) {
			char[][] map = new char[n][];
			for (int i = 0; i < n; i++)
				map[i] = ns(m);
			return map;
		}
	 
		private int[] na(int n) {
			int[] a = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = ni();
			return a;
		}
		private long[] nl(int n) {
			long[] a = new long[n];
			for (int i = 0; i < n; i++)
				a[i] = nl();
			return a;
		}
	 
		private int ni() {
			int num = 0, b;
			boolean minus = false;
			while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
				;
			if (b == '-') {
				minus = true;
				b = readByte();
			}
	 
			while (true) {
				if (b >= '0' && b <= '9') {
					num = num * 10 + (b - '0');
				} else {
					return minus ? -num : num;
				}
				b = readByte();
			}
		}
	 
		private long nl() {
			long num = 0;
			int b;
			boolean minus = false;
			while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
				;
			if (b == '-') {
				minus = true;
				b = readByte();
			}
	 
			while (true) {
				if (b >= '0' && b <= '9') {
					num = num * 10 + (b - '0');
				} else {
					return minus ? -num : num;
				}
				b = readByte();
			}
		}
	 
}