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:

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<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void IO() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
vector<pair<int, int>> ans;
int n, d;
bool dfs(int cnt, int sum, int dep, int limit) {
  if (cnt * dep > sum) return false;
  if (cnt == 0 && sum == 0) return true;
  for (int i = 1; i <= min(limit, cnt); i++) {
    if (i * dep > sum) return false;
    if (dfs(cnt - i, sum - i * dep, dep + 1, i * 2)) {
      ans.emplace_back(dep, i);
      return true;
    }
  }
  return false;
}
void solve() {
  ans.clear();
  int mx = 0, mn = 0;
  cin >> n >> d;
  mx = n * (n - 1) / 2;
  for (int i = 0, tem = 1, nn = n; nn; i++, nn -= min(nn, tem), tem *= 2)
    mn += i * min(tem, nn);
  if (dfs(n - 1, d, 1, 2)) {
    cout << "YES" << '\n';
    reverse(ans.begin(), ans.end());
    vector<int> pre(1, 1), pp;
    int id = 1;
    for (auto x : ans) {
      pp.clear();
      for (int i = 0; i < x.second; i++)
        cout << pre[i / 2] << ' ', pp.push_back(++id);
      pre = pp;
    }
    cout << '\n';
  } else
    cout << "NO" << '\n';
}
int main() {
  IO();
  int _;
  cin >> _;
  while (_--) solve();
}