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>
#pragma GCC optimize("O3")
using namespace std;
const double EPS = 1e-9;
const int N = 5e5 + 9;
set<int> a[5005];
int pr[5005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n, d;
    cin >> n >> d;
    for (int i = 1; i <= n; i++) a[i].clear(), pr[i] = i;
    if (d > n * (n - 1) / 2) {
      cout << "NO" << endl;
      continue;
    }
    set<pair<int, int>> k;
    int x = n * (n - 1) / 2;
    for (int i = 1; i <= n - 1; i++)
      a[i].insert(i + 1), k.insert({i, i}), pr[i] = max(1, i - 1);
    k.insert({n, n}), pr[n] = max(1, n - 1);
    while (x > d) {
      auto it = --k.end();
      auto u = *it;
      k.erase(it);
      a[pr[u.second]].erase(u.second);
      int mov = x - d;
      auto cat = k.lower_bound({u.first - mov - 1, -1e9});
      auto v = *cat;
      if (u.first - v.first - 1 <= 0) break;
      x -= u.first - v.first - 1;
      a[v.second].insert(u.second);
      if (a[v.second].size() == 2) k.erase(cat);
      k.insert({v.first + 1, u.second});
      pr[u.second] = v.second;
    }
    if (x == d) {
      cout << "YES" << endl;
      for (int i = 2; i <= n; i++) cout << pr[i] << " ";
      cout << endl;
    } else
      cout << "NO" << endl;
  }
  return 0;
}