1311_E. Construct the Binary Tree
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>
```python
t=int(input())
import math as m
def print_tree(l):
    t=[{'p':None, 'c':[]} for i in range(sum(l))]
    l2={i:[] for i in range(len(l))}
    j=0
    for i in range(sum(l)):
        l2[j].append(i+1)
        if len(l2[j])==l[j]:
            j+=1
    for i in range(1,len(l)):
        p=0
        for n in l2[i]:
            pi=l2[i-1][p]
            t[n-1]['p']=pi
            t[pi-1]['c'].append(n)
            if len(t[pi-1]['c'])==2:
                p+=1

    for i in range(1, len(t)):
        print(t[i]['p'], end=' ')
    print()
for _ in range(t):
    n, d = map(int, input().split())
    k=m.floor(m.log2(n))
    maxd=n*(n-1)//2
    mind=k*2**(k+1)-2**(k+1)+2+k*(n-2**(k+1)+1)
    if (d<mind) or (d>maxd):
        print("NO")
        continue
    l=[1 for i in range(n)]
    s=maxd
    lowest=1
    highest=n-1
    while (s>d):
        diff=s-d
        if (diff>highest-lowest):
            l[highest]-=1
            l[lowest]+=1
            s-=(highest-lowest)
            highest-=1
            if l[lowest-1]*2==l[lowest]:
                lowest+=1
        else:
            level=highest-diff
            l[level]+=1
            l[highest]-=1
            s-=diff
    print("YES")
    print_tree(l)
    

```