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>
for _ in range(int(input())):
    n, d = map(int, input().split())
    ld, rd = 0, n * (n - 1) // 2
    tn = n
    cd = 0
    mvl = 1
    while tn:
        vl = min(tn, mvl)
        ld += cd * vl
        cd += 1
        tn -= vl
        mvl *= 2
    if not (ld <= d <= rd):
        print('NO')
        continue
    parent = [i-1 for i in range(n)]
    bad = [False] * n
    deep = [i for i in range(n)]
    child = [1] * n
    child[-1] = 0
    cur = rd
    while cur > d:
        v = None
        for i in range(n):
            if not bad[i] and not child[i] and (v is None or deep[i] < deep[v]):
                v = i
        p = None
        for i in range(n):
            if child[i] < 2 and deep[i] == deep[v] - 2 and (p is None or deep[i] > deep[p]):
                p = i
        if p is None:
            bad[v] = True
        else:
            child[parent[v]] -= 1
            deep[v] -= 1
            child[p] += 1
            parent[v] = p
            cur -= 1
    print('YES')
    print(*map(lambda x: x + 1, parent[1:]))
