You are given two integers n and d. You need to construct a rooted binary tree consisting of n vertices with a root at the vertex 1 and the sum of depths of all vertices equals to d.

A tree is a connected graph without cycles. A rooted tree has a special vertex called the root. A parent of a vertex v is the last different from v vertex on the path from the root to the vertex v. The depth of the vertex v is the length of the path from the root to the vertex v. Children of vertex v are all vertices for which v is the parent. The binary tree is such a tree that no vertex has more than 2 children.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.

The only line of each test case contains two integers n and d (2 ≤ n, d ≤ 5000) — the number of vertices in the tree and the required sum of depths of all vertices.

It is guaranteed that the sum of n and the sum of d both does not exceed 5000 (∑ n ≤ 5000, ∑ d ≤ 5000).

Output

For each test case, print the answer.

If it is impossible to construct such a tree, print "NO" (without quotes) in the first line. Otherwise, print "{YES}" in the first line. Then print n-1 integers p_2, p_3, ..., p_n in the second line, where p_i is the parent of the vertex i. Note that the sequence of parents you print should describe some binary tree.

Example

Input


3
5 7
10 19
10 18


Output


YES
1 2 1 3 
YES
1 2 3 3 9 9 2 1 6 
NO

Note

Pictures corresponding to the first and the second test cases of the example:

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import java.io.*;
import java.util.*;
 
public class Main {

	public static void main(String[] args)throws IOException {
        InputReader in = new InputReader(System.in);
        PrintWriter out = new PrintWriter(System.out);
        
		int t = in.nextInt();
		int n,d,lvl,mx,ix,ct;
		long val;
		Deque deques[];
		int par[];
		for(int x=0;x<t;x++)
		{
			n=in.nextInt();
			d=in.nextInt();
			ct=0;
			deques = new LinkedList[n];
			par = new int[n];
			for(int j=0;j<n;j++)
				deques[j]=new LinkedList<Integer>();
			deques[0].addLast(0);
			lvl=1; ct=2;
			val=0;
			for(int j=1;j<n;j++)
			{
				if(ct==0)
				{
					lvl++;
					ct=(int)Math.pow(2,lvl);
				}
				deques[lvl].addLast(j);
				val+=(lvl);
				ct--;
			}
			// out.println("val="+val);
			if(d<val)
			{
				out.println("NO");
				continue;
			}
			while(true)
			{
				if(val==d)
				{
					break;
				}
				ix=-1;
				for(int i=n-1;i>=1;i--)
				{
					if(deques[i].size()<=1)
						continue;
					ix=i;
					break;
				}
				// out.println("ix="+ix);
				if(ix==-1)
					break;
				deques[ix+1].addLast(deques[ix].pollLast());
				val++;
			}
			// out.println("val="+val);
			if(val!=d)
			{
				out.println("NO");
				continue;
			}
			out.println("YES");
			ct=0;
			for(int i=0;i<n;i++)
				par[i]=-1;
			for(int i=1;i<n;i++)
			{
				ct=0;
				if(deques[i].size()==0)
					continue;
				val = (int)deques[i-1].pollFirst()+1;
				for(int j=0;j<deques[i].size();j++)
				{
					if(ct==2)
					{
						val=(int)deques[i-1].pollFirst()+1;
						ct=0;
					}
					mx=(int)deques[i].pollFirst();
					par[mx-1]=(int)val;
					deques[i].addLast(mx);
					// out.print(val+" ");
					ct++;
				}	
			}
			for(int i=0;i<n;i++){
				if(par[i]!=-1)
					out.print(par[i]+" ");
			}
			out.println();
		}

        out.close();
	}
    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

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

    }

}