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 CF1311E extends PrintWriter {
	CF1311E() { super(System.out); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1311E o = new CF1311E(); o.main(); o.flush();
	}

	void main() {
		int t = sc.nextInt();
		while (t-- > 0) {
			int n = sc.nextInt();
			int s = sc.nextInt();
			int[] kk = new int[n];
			if (s > n * (n - 1) / 2) {
				println("NO");
				continue;
			}
			for (int d = 0, k = n; d < n && k > 0; d++)
				while (k > 0 && kk[d] < 1 << d) {
					k--; kk[d]++;
					s -= d;
				}
			if (s < 0) {
				println("NO");
				continue;
			}
			println("YES");
			while (s > 0)
				for (int d = 1; d + 1 < n && s > 0; d++)
					while (s > 0 && kk[d + 1] + 1 <= (kk[d] - 1) * 2) {
						s--; kk[d + 1]++; kk[d]--;
					}
			for (int d = 1, i = 1, j = 2; d < n; d++) {
				int i_ = j;
				while (kk[d]-- > 0) {
					print(i + " "); j++;
					if (kk[d]-- > 0) {
						print(i + " "); j++;
					}
					i++;
				}
				i = i_;
			}
			println();
		}
	}
}