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())
for T in range(t):
    n, d = map(int, input().strip().split())
    u = (n-1) * n // 2
    if d > u: print("NO"); continue
    l = t = 0
    for i in range(1, 1+n):
        l += t
        if not i & i+1: t += 1
    if d < l: print("NO"); continue
    p = [1] * n
    l = [1 << i for i in range(n)]
    o = n-1
    while u > d:
        m = o
        while u > d and m and p[m-1] < l[m-1]: m -= 1; u -= 1;
        p[o] -= 1;
        p[m] += 1;
        o -= 1
    r = [0] * n
    c = [1, 1]
    t = []
    v = 1
    for i in range(1, n):
        if not p[i]: break
        for j in range(p[i]):
            r[v] = c.pop()
            v += 1
            t.append(v)
            t.append(v)
        c = t
        t = []
    print("YES")
    print(*(r[i] for i in range(1, n)))
