Problem: 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:

<image>

<image>

Solution (in JAVA):

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

public class TestClass {

	static PrintWriter out = new PrintWriter(System.out);
	public static void main(String[] args) throws IOException{
 
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		int t = Integer.parseInt(in.readLine());
loop:		for(int ti=0;ti<t;ti++) {
			String s1[] = in.readLine().split(" ");
			int n = Integer.parseInt(s1[0]);
			int depth = Integer.parseInt(s1[1]);
			int maxd = n*(n-1)/2;
			int mind = 0, cnt = 0;
			for(int i=2;i<=n;i++) {
				if(i!=0 && (i & (i-1))==0) {
					cnt++;
				}
				mind += cnt;
			}
			if(!(depth<=maxd && depth>=mind)) {
				out.println("NO");
				continue loop;
			}
			int par[] = new int[n];
			for(int i=1;i<n;i++) {
				par[i] = i-1;
			}
			int cntchild[] = new int[n];
			for(int i=0;i<n-1;i++) {
				cntchild[i] = 1;
			}
			int dep[] = new int[n];
			for(int i=1;i<n;i++) {
				dep[i] = i;
			}
			int d = maxd;
			int v[] = new int[n];
			while(d!=depth) {
				int topleaf = -1;
				for(int i=0;i<n;i++) {
					if(cntchild[i]==0 && v[i]==0) {
						if(topleaf==-1) {
							topleaf=i;
						}
						else if(dep[i]<dep[topleaf]) {
							topleaf = i;
						}
					}
				}
				int newp = -1;
				for(int i=0;i<n;i++) {
					//System.out.println(topleaf);
					if(cntchild[i]<2 && dep[i]==dep[topleaf]-2) {
						newp = i;
						break;
					}
				}
				if(newp==-1) {
					v[topleaf] = 1;
					continue;
				}
				d--;
				cntchild[par[topleaf]]--;
				cntchild[newp]++;
				par[topleaf] = newp;
				dep[topleaf] = dep[newp]+1;
			}
			out.println("YES");
			for(int i=1;i<n;i++) {
				out.print(par[i]+1+" ");
			}
			out.println();
		}
        out.close();
	}	
	
}