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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def solve(n, d):
    arr  = [[x] for x in range(1, n+1)]
    S    = sum(range(1, n))

    cur  = 1 
    while S > d and cur != len(arr):
        need = S-d
        can  = len(arr) - 1 - cur
    
        if can < need:
            S -= can
            arr[cur] += arr[-1]
            arr.pop()
        else:
            S -= need
            arr[len(arr)-1-need] += arr[-1]
            arr.pop()
        
        if cur == len(arr):
            break  
        elif len(arr[cur]) == 2**cur:
            cur+=1
        
    if S != d:
        print('NO')
    else:
        print('YES')
        ans = [0] * (n-1)
        
        for i, a in enumerate(arr):
            if i == 0:
                continue
            
            for j, v in enumerate(a):
                ans[v-2] = str(arr[i-1][j//2])  
        print(' '.join(ans))        
        
for _ in range(int(input())):
    n, d = map(int, input().split())
    solve(n, d)