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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<image>
from math import ceil,floor,log
t=int(input())
for i2 in range(t):
    n,d=map(int,input().split())
    p,dad=0,[0]*(n+1)
    c=(n*(n-1))//2
    for i in range(1,n+1):
        p+=floor(log(i,2))
    if p>d or c<d:
        print("NO")
        continue
    now=1
    dad=[0]*(n+1)
    vec=[[]for i in range(n+1)]
    for i in range(1,n+1):
        vec[i-1].append(i)
    for i in range(n,0,-1):
        if c-d<=i-1-now:
            vec[now].append(i)
            dad[i]=vec[i-2-(c-d)][(len(vec[i-(c-d)-1])+1)//2-1]
            break
        else:
            vec[now].append(i)
            dad[i]=vec[now-1][(len(vec[now])+1)//2-1]
            c-=i-1-now
            if len(vec[now])>=2**now:
                now+=1
    print("YES")
    for i in range(2,n+1):
        if dad[i]==0:
            print(i-1,end=" ")
        else:
            print(dad[i],end=" ")
    print("")