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>
import java.io.*;
import java.util.*;

@SuppressWarnings("unchecked")
public class Main {
	public static void main(String[] args) {
		InputReader in = new InputReader();
		StringBuilder out = new StringBuilder();

		int T = in.nextInt();
		while (T-- > 0) {
			int N = in.nextInt();
			int D = in.nextInt();
			int lb, x;
			lb = x = 0;
			for (int i = 2; i <= N; i++) {
				if ((i & (i - 1)) == 0) {
					x++;
				}
				lb += x;
			}
			if (D < lb || N * (N - 1) / 2 < D) {
				out.append("NO\n");
				continue;
			}
			int[] par = new int[N + 1];
			int[] deg = new int[N + 1];
			List<Integer>[] dep = new List[N];
			for (int i = 0; i < N; i++) {
				dep[i] = new ArrayList<>();
				dep[i].add(i + 1);
			}
			for (int i = 2; i <= N; i++) {
				par[i] = i - 1;
			}
			Arrays.fill(deg, 1);
			deg[N] = 0;
			int now, cur_depth;
			now = cur_depth = -1;
			int sum = N * (N - 1) / 2;
			while (sum > D) {
				for (int i = N - 1; now == -1; i--) {
					Iterator<Integer> it = dep[i].iterator();
					while (it.hasNext()) {
						x = it.next();
						if (deg[x] == 0) {
							now = x;
							cur_depth = i;
							break;
						}
					}
				}
				boolean flag = false;
				dep[cur_depth].remove((Integer)now);
				cur_depth--;
				dep[cur_depth].add(now);
				for (int y : dep[cur_depth - 1]) {
					if (deg[y] < 2) {
						deg[par[now]]--;
						par[now] = y;
						deg[y]++;
						flag = true;
						break;
					}
				}
				if (flag) {
					sum--;
				} else {
					dep[cur_depth].remove((Integer)now);
					dep[cur_depth + 1].add(now);
					now = -1;
				}
				if (cur_depth == 1) {
					now = -1;
				}
			}
			out.append("YES\n");
			for (int i = 2; i <= N; i++) {
				out.append(par[i]).append(' ');
			}
			out.setCharAt(out.length() - 1, '\n');
		}

		System.out.print(out);
	}

	static class InputReader {
		public BufferedReader reader;
		public StringTokenizer st;

		public InputReader() {
			reader = new BufferedReader(new InputStreamReader(System.in));
		}

		public String next() {
			while (st == null || !st.hasMoreTokens()) {
				st = new StringTokenizer(nextLine());
			}
			return st.nextToken();
		}

		public String nextLine() {
			try {
				return reader.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return null;
		}

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