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;
queue<int> able[5010];
int fa[5110], lft[5110];
signed main() {
  std::ios::sync_with_stdio(false);
  int T;
  cin >> T;
  while (T--) {
    int n, d;
    cin >> n >> d;
    int now = n * (n - 1) / 2;
    if (now < d) {
      cout << "NO\n";
      continue;
    }
    for (int i = 1; i <= n; i += 1) {
      able[i] = queue<int>();
      able[i].push(i);
      lft[i] = 1;
      fa[i] = i - 1;
    }
    ++lft[n];
    int l = 1, endfg = 0, flag = 0;
    for (int i = n;; i -= 1) {
      while (1) {
        while (able[l].size() && lft[able[l].front()] == 0) able[l].pop();
        if (able[l].empty())
          ++l;
        else
          break;
      }
      if (now == d) break;
      if (now - d > i - 1 - l)
        ;
      else
        l = i - 1 - now + d, endfg = 1;
      if (i - 1 <= l) {
        flag = 1;
        break;
      } else {
        ++lft[fa[i]];
        now -= i - 1 - l;
        fa[i] = able[l].front();
        --lft[able[l].front()];
        able[l + 1].push(i);
      }
    }
    if (flag) {
      cout << "NO" << '\n';
    } else {
      cout << "YES\n";
      for (int i = 2; i <= n; i += 1) {
        cout << fa[i] << ' ';
      }
      cout << '\n';
    }
  }
  return 0;
}