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:
import sys
from array import array  # noqa: F401
from typing import List, Tuple, TypeVar, Generic, Sequence, Union  # noqa: F401


def input():
    return sys.stdin.buffer.readline().decode('utf-8')


def main():
    t = int(input())
    for _ in range(t):
        n, d = map(int, input().split())
        depth = [0] + list(range(n))
        child = [0] + [1] * (n - 1) + [0]
        par = [-1, -1] + list(range(1, n))
        cur_d = sum(depth)

        if d > cur_d:
            print('NO')
            continue

        for i in range(n, 0, -1):
            if child[i]:
                continue
            child[par[i]] -= 1
            p, delta = -1, 0

            for j in range(1, n + 1):
                if i != j and child[j] < 2 and cur_d - d >= depth[i] - depth[j] - 1 and depth[i] - depth[j] - 1 > delta:
                    delta = depth[i] - depth[j] - 1
                    p = j
            if p != -1:
                child[p] += 1
                par[i] = p
                depth[i] = depth[p] + 1
                cur_d -= delta
            else:
                break

        if cur_d == d:
            print('YES')
            print(*par[2:])
        else:
            print('NO')


if __name__ == '__main__':
    main()