Problem: Kuroni is the coordinator of the next Mathforces round written by the "Proof by AC" team. All the preparation has been done, and he is discussing with the team about the score distribution for the round.

The round consists of n problems, numbered from 1 to n. The problems are ordered in increasing order of difficulty, no two problems have the same difficulty. A score distribution for the round can be denoted by an array a_1, a_2, ..., a_n, where a_i is the score of i-th problem. 

Kuroni thinks that the score distribution should satisfy the following requirements:

  * The score of each problem should be a positive integer not exceeding 10^9. 
  * A harder problem should grant a strictly higher score than an easier problem. In other words, 1 ≤ a_1 < a_2 < ... < a_n ≤ 10^9. 
  * The balance of the score distribution, defined as the number of triples (i, j, k) such that 1 ≤ i < j < k ≤ n and a_i + a_j = a_k, should be exactly m. 



Help the team find a score distribution that satisfies Kuroni's requirement. In case such a score distribution does not exist, output -1.

Input

The first and single line contains two integers n and m (1 ≤ n ≤ 5000, 0 ≤ m ≤ 10^9) — the number of problems and the required balance.

Output

If there is no solution, print a single integer -1.

Otherwise, print a line containing n integers a_1, a_2, ..., a_n, representing a score distribution that satisfies all the requirements. If there are multiple answers, print any of them.

Examples

Input


5 3


Output


4 5 9 13 18

Input


8 0


Output


10 11 12 13 14 15 16 17


Input


4 10


Output


-1

Note

In the first example, there are 3 triples (i, j, k) that contribute to the balance of the score distribution. 

  * (1, 2, 3) 
  * (1, 3, 4) 
  * (2, 4, 5) 

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;

public class Main {
	public static void main(String args[]) {new Main().run();}

	FastReader in = new FastReader();
	PrintWriter out = new PrintWriter(System.out);
	void run(){
		work();
		out.flush();
	}
	long mod=1000000007;
	long gcd(long a,long b) {
		return a==0?b:gcd(b%a,a);
	}
	void work() {
		int n=ni(),m=ni();
		if(n<=2) {
			if(n<=2&&m>0) {
				out.println(-1);
				return;
			}
			for(int i=1;i<=n;i++) {
				out.print(i+" ");
			}
			return;
		}
		ArrayList<Integer> ret=new ArrayList<>();
		ret.add(1);
		ret.add(2);
		int sum=0;
		int i=3;
		for(;i<=n;i++) {
			if(sum+(i-1)/2>=m) {
				int r=m-sum;
				int v1=i-1;
				//(v1-v2+1)/2==r v2=v1+1-2*r
				int v2=v1+1-2*r;
				ret.add(v1+v2);
				sum=m;
				i++;
				break;
			}else {
				sum+=(i-1)/2;
				ret.add(i);
			}
		}
		if(sum<m) {
			out.println(-1);
		}else {
			for(int cur=10000001;i<=n;i++) {
				cur+=10000;
				ret.add(cur);
			}
			for(int r:ret) {
				out.print(r+" ");
			}
		}
	}


	
	//input
	@SuppressWarnings("unused")
	private ArrayList<Integer>[] ng(int n, int m) {
		ArrayList<Integer>[] graph=(ArrayList<Integer>[])new ArrayList[n];
		for(int i=0;i<n;i++) {
			graph[i]=new ArrayList<>();
		}
		for(int i=1;i<=m;i++) {
			int s=in.nextInt()-1,e=in.nextInt()-1;
			graph[s].add(e);
			graph[e].add(s);
		}
		return graph;
	}

	private ArrayList<long[]>[] ngw(int n, int m) {
		ArrayList<long[]>[] graph=(ArrayList<long[]>[])new ArrayList[n];
		for(int i=0;i<n;i++) {
			graph[i]=new ArrayList<>();
		}
		for(int i=1;i<=m;i++) {
			long s=in.nextLong()-1,e=in.nextLong()-1,w=in.nextLong();
			graph[(int)s].add(new long[] {e,w,i});
			graph[(int)e].add(new long[] {s,w});
		}
		return graph;
	}

	private int ni() {
		return in.nextInt();
	}

	private long nl() {
		return in.nextLong();
	}
	
	private String ns() {
		return in.next();
	}

	private long[] na(int n) {
		long[] A=new long[n];
		for(int i=0;i<n;i++) {
			A[i]=in.nextLong();
		}
		return A;
	}
	private int[] nia(int n) {
		int[] A=new int[n];
		for(int i=0;i<n;i++) {
			A[i]=in.nextInt();
		}
		return A;
	}
}	

class FastReader
{
	BufferedReader br;
	StringTokenizer st;

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


	public String next() 
	{
		while(st==null || !st.hasMoreElements())//回车，空行情况
		{
			try {
				st = new StringTokenizer(br.readLine());
			} catch (IOException e) {
				e.printStackTrace();
			}
		}
		return st.nextToken();
	}

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

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