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>
import sys
range = xrange
input = raw_input

inp = [int(x) for x in sys.stdin.read().split()]; ii = 0

t = inp[ii]; ii += 1
for _ in range(t):
    n = inp[ii]; ii += 1
    d = inp[ii]; ii += 1

    if d > n * (n - 1)//2:
        print 'NO'
        continue


    summa = 0
    layers = [[] for _ in range(n)]
    m = n
    bit = 0
    used = 0
    while m:
        dm = min(m, 1 << bit)
        summa += dm * bit
        m -= dm
        for _ in range(dm):
            layers[bit].append(used)
            used += 1
        bit += 1
    
    if d < summa:
        print 'NO'
        continue

    for maxused in reversed(range(len(layers))):
        if layers[maxused]:
            break
    layer = len(layers) - 1
    while True:
        while summa + maxused + 1 - layer <= d and len(layers[layer]) > 1:
            i = layers[layer].pop()
            summa += maxused + 1 - layer
            maxused += 1
            layers[maxused].append(i)

        if layer > 0 and len(layers[layer]) <= 1:
            layer -= 1
        else:
            break
    if summa != d:
        i = layers[layer].pop()
        layers[layer + d - summa].append(i)
    
    used = 0
    for layer in range(len(layers)):
        l = len(layers[layer])
        layers[layer] = list(range(used, used + l))
        used += l

    print 'YES'
    P = [-1]*n
    for layer in range(1, len(layers)):
        for i in range(len(layers[layer])):
            P[layers[layer][i]] = layers[layer - 1][i >> 1]
      
    
    score_calc = 0
    for layer in range(len(layers)):
        score_calc += len(layers[layer]) * layer
    assert d == score_calc

    print ' '.join(str(x + 1) for x in P[1:])
