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.util.Arrays;
import java.util.PriorityQueue;
import java.util.Scanner;

public class Test2 {
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);

        int t = s.nextInt();

        while(t-- > 0){
            int n = s.nextInt();
            int d = s.nextInt();

            int maxD = (n*(n-1))/2;
            int minD = 0;
            int cntD = 0;
            for(int i=2;i<=n;i++){
                if((i & (-i)) == i){
                    cntD++;
                }
                minD += cntD;
            }
            if(!(minD <= d && d<=maxD)){
                System.out.println("NO");
                continue;
            }

            int[] cntChild = new int[n];
            Arrays.fill(cntChild, 1);
            cntChild[n-1] = 0;

            int[] par = new int[n];
            par[0] = -1;
            for(int i=1;i<n;i++){
                par[i] = i-1;
            }

            int[] dep = new int[n];
            for(int i=0;i<n;i++){
                dep[i] = i;
            }

            boolean[] bad = new boolean[n];

            int curr = maxD;
            boolean fault = false;

            while(curr>d){
                int v = -1;
                for(int i=0;i<n;i++){
                    if(!bad[i] && cntChild[i] == 0){
                        if(v == -1)
                            v = i;
                        else if(dep[i] < dep[v]){
                            v = i;
                        }
                    }
                }

                int p = -1;
                for(int i=0;i<n;i++){
                    if(cntChild[i] < 2 && dep[i] == dep[v] - 2){
                        if(p == -1)
                            p = i;
                        else if(dep[i] > dep[p])
                            p = i;
                    }
                }
                if(p == -1){
                    bad[v] = true;
                    continue;
                }

                cntChild[par[v]]--;
                cntChild[p]++;
                par[v] = p;
                dep[v]--;
                curr--;
            }


            System.out.println("YES");
            for(int i=1;i<n;i++){
                System.out.print((par[i] + 1) + " ");
            }
            System.out.println();

        }
    }
}