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>
import bisect
import sys
input = sys.stdin.readline

t = int(input())
pow2 = [1]
for _ in range(20):
    pow2.append(2 * pow2[-1])
pow2 = set(pow2)
for _ in range(t):
    n, d = map(int, input().split())
    p = []
    d0 = []
    now = 0
    l, l0 = [1], [1]
    for i in range(2, n + 1):
        p.append(i // 2)
        if i in pow2:
            l.append(i)
            l0.append(i)
            now += 1
        d0.append(now)
    s = sum(d0)
    ans = "YES" if s <= d <= n * (n - 1) // 2 else "NO"
    print(ans)
    if ans == "NO":
        continue
    c = d - s
    for i in range(n, 0, -1):
        if not c:
            break
        if i in pow2:
            continue
        j = bisect.bisect_left(l, i) - 1
        for k in range(j, len(l0) - 1):
            if not c:
                break
            c -= 1
            p[i - 2] = l0[k]
            d0[i - 2] += 1
        if not c:
            break
        c -= 1
        p[i - 2] = l0[-1]
        d0[i - 2] += 1
        l0.append(i)
    print(*p)