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 ConstructTheBinaryTree {
    public static void main(String[] args) throws IOException{
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        int t = Integer.parseInt(f.readLine());
        for(int x = 0; x < t; x++){
            StringTokenizer st = new StringTokenizer(f.readLine());
            int n = Integer.parseInt(st.nextToken());
            int d = Integer.parseInt(st.nextToken());
            long sum = 0;
            long lower_bound = 0;
            int[] par = new int[n+1];
            int[] depths = new int[n+1];
            TreeSet<Integer>[] pos = new TreeSet[n+1];
            int[] children = new int[n+1];
            for(int i = 1; i <= n; i++){
                sum+=(i-1);
                par[i] = i-1;
                depths[i] = i-1;
                pos[i] = new TreeSet<Integer>();
                pos[i].add(i);
                if(i != n) children[i] = 1;
                else children[i] = 0;
            }
            if(d > sum){
                out.println("NO");
                continue;
            }
            long temp = n;
            long idx = 0;
            while(true){
                long amt = (1L << idx)*idx;
                if((1L << idx) >= temp){
                    lower_bound+=(temp)*idx;
                    break;
                }
                lower_bound+=amt;
                temp-=(1L << idx);
                idx++;
            }
            if(d < lower_bound){
                out.println("NO");
                continue;
            }
            if(d == sum){
                out.println("YES");
                StringBuilder ans = new StringBuilder();
                for(int i = 2; i <= n; i++){
                    ans.append(par[i]);
                    ans.append(" ");
                }
                out.println(ans.toString().trim());
                continue;
            }
            for(int i = n; i >= 0; i--){
                boolean found = false;
                int temp2 = i;
                while(temp2 > 2 && !pos[temp2-2].isEmpty()) {
                    children[par[i]]--;
                    if(children[par[i]] > 0) pos[temp2-1].add(par[i]);
                    pos[temp2].remove(i);
                    sum--;
                    temp2--;
                    par[i] = pos[temp2-1].first();
                    children[par[i]]++;
                    if(children[par[i]] == 2) pos[temp2-1].remove(par[i]);
                    pos[temp2].add(i);
                    if(sum == d){
                        found = true;
                        break;
                    }
                }
                if(found){
                    break;
                }
            }
            out.println("YES");
            StringBuilder ans = new StringBuilder();
            for(int i = 2; i <= n; i++){
                ans.append(par[i]);
                ans.append(" ");
            }
            out.println(ans.toString().trim());
        }
        out.close();
    }
}