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.util.Scanner;
import java.util.ArrayList;

public class R624_E {
	private static void solve (int n, int d) {
		int upperBound = n*(n-1) / 2;
		int low = 0;
		for (int i = 1, cd = 0; i <= n; i++) {
			if ((i & (i-1)) == 0) cd++;
			low += cd - 1;
		}
		if (d < low || d > upperBound) {
			System.out.println("NO");
			return;
		}
		
		int[] dep = new int[n];
		int[] par = new int[n];
		int[] bad = new int[n];
		int[] children = new int[n];
		int sum = upperBound;

		for (int i = 1; i < n; i++) {
			children[i-1]++;
			dep[i] = dep[i-1]+1;
			par[i] = i-1;
		}	
		while (sum != d) {
			int v = -1;
			int p = -1;
			for (int i = 0; i < n; i++) {
				if (bad[i] == 0 && children[i] == 0 && (v== -1 || dep[v] > dep[i])) {
					v = i;
				}
			}
			if (v == -1) {
				System.out.println("NO");
				return;
			}
			for (int i = 0; i < n; i++) {
				if (children[i] < 2 && dep[i] < dep[v] - 1 && (p==-1 || dep[p] < dep[i]) ) {
					p = i;
				}
			}
			
			if (p != -1) {
				children[p]++;
				sum--;
				children[par[v]]--;
				par[v] = p;
				dep[v] = dep[p]+1;
			} else {
				bad[v] = 1;
			}
		}
		
		System.out.println("YES");
			printout(par);
	}
	private static void printout(int[] par) {
		for (int i = 1; i < par.length; i++) {
			System.out.print(par[i] + 1 + " ");
		}
		System.out.println();
	}
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int t = in.nextInt();
		for (int q = 0; q < t; q++) {
			int n = in.nextInt();
			int d = in.nextInt();
			solve(n, d);
		}
		in.close();
	}
}
