Problem: 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>

Solution (in PYTHON3):

Solution:

t = int(input())

for loop in range(t):

    n,d = map(int,input().split())

    lis = [1] * n
    nsum = 0
    for i in range(n):
        nsum += i

    sub = nsum - d

    if d > nsum:
        print ("NO")
        continue

    ableind = 1
    pickind = n-1

    while pickind > ableind:

        if sub > pickind - ableind:
            lis[pickind] -= 1
            lis[ableind] += 1
            sub -= pickind - ableind
        else:
            lis[pickind] -= 1
            lis[pickind - sub] += 1
            sub = 0
            break

        if lis[ableind] == 2**ableind:
            ableind += 1
        if lis[pickind] == 0:
            pickind -= 1

    if sub > 0:
        print ("NO")
    else:
        print ("YES")
        ans = []

        for i in range(n):
            if i == 0:
                ans.append([1])
                ind = 2
            else:
                nlis = []
                for j in range(lis[i]):
                    nlis.append(ind)
                    ind += 1
                ans.append(nlis)

        pl = []

        for i in range(1,n):
            for j in range(lis[i]):
                pl.append(ans[i-1][j//2])

        print (" ".join(map(str,pl)))