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>
t=int(input())
pows=[2**i for i in range(21)]
for _ in range(t):
  n,d=map(int,input().split())
  mins=0
  for k in range(20):
    if pows[k+1]<=n:
      mins+=k*pows[k]
    else:
      mins+=k*(n-(pows[k]-1))
      break
  maxs=n*(n-1)//2
  if d<mins or d>maxs:
    print('NO')
  else:
    print('YES')
    ans=[i for i in range(n)]
    cnt=1
    base=1
    for i in range(n-1,-1,-1):
      if maxs-(ans[i]-base)>d:
        maxs-=ans[i]-base
        ans[i]=base
        cnt+=1
        if cnt==pows[base]:
          base+=1
          cnt=1
      else:
        ans[i]-=(maxs-d)
        break
    ans=sorted(ans)
    arr=[[] for _ in range(n+1)]
    for i in range(n):
      arr[ans[i]].append(i+1)
    ret=[]
    for i in range(1,n+1):
      pos=0
      cnt=0
      for j in range(len(arr[i])):
        ret.append(arr[i-1][pos])
        cnt+=1
        if cnt==2:
          pos+=1
          cnt=0
    print(*ret)