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:

<image>

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
namespace Acc {
const int N = 5e3 + 10;
int fa[N], d[N], a[N], n, D, b[N], mx;
void work() {
  cin >> n >> D;
  if (D > n * (n - 1) / 2) return puts("NO"), void(0);
  for (int i = 2; i <= n; i++)
    fa[i] = i / 2, d[i] = d[fa[i]] + 1, D -= d[i], b[i] = a[i] = 0;
  if (mx = d[n], D < 0) return puts("NO"), void(0);
  for (int i = n; i; i >>= 1) b[i] = 1, a[d[i]] = i;
  for (int i = n, las; las = mx, i; i--)
    if (!b[i])
      while (D && d[fa[i]] < las) {
        fa[i] = a[d[i]], d[i] = d[fa[i]] + 1, D--;
        if (d[i] > mx) mx++, a[d[i]] = i, b[i] = 1;
      }
  if (D) puts("NO"), void(0);
  puts("YES");
  for (int i = 2; i <= n; i++) cout << fa[i] << ' ';
  puts("");
}
}  // namespace Acc
int main() {
  int T;
  for (cin >> T; T--;) Acc::work();
  return 0;
}