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:
for tests in range(int(input())):
    n, d = map(int,input().split())
    upper_bound, lower_bound, lv = n * (n - 1) // 2, 0, 0
    for i in range(1, n + 1):
        if not(i & (i - 1)):
            lv += 1
        lower_bound += lv - 1
    if not(d in range(lower_bound, upper_bound + 1)):
        print("NO")
    else:
        bad = [False] * n
        depth = list(range(n))
        count_child = [1] * n
        count_child[n - 1] = 0
        current_depth = upper_bound
        parent = list(range(-1,n-1))
        while current_depth > d:
            v, p = -1, -1
            for i in range(n):
                if not(bad[i]) and count_child[i] == 0 and (v == -1 or depth[i] < depth[v]):
                    v = i
            for i in range(n):
                if count_child[i] < 2 and depth[i] == depth[v] - 2 and (p == -1 or depth[i] > depth[p]):
                    p = i
            if p == -1:
                bad[v] = True
                continue
            count_child[parent[v]] -= 1
            depth[v] -= 1
            count_child[p] += 1
            parent[v] = p
            current_depth -= 1
        print("YES")
        print(*((parent[i] + 1) for i in range(1,n)))