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;
const int mod = 1e9 + 7;
const int N = 5e5 + 10;
void solve() {
  long long n, d;
  cin >> n >> d;
  if (n * (n - 1) / 2 < d) {
    cout << "NO\n";
    return;
  }
  long long dep = (n * (n - 1) / 2);
  set<int> leaf, av;
  vector<int> par(n + 1), lev(n + 1), ch(n + 1);
  for (int i = 2; i <= n; i++) {
    par[i] = i - 1;
    lev[i] = i - 1;
    av.insert(i);
    ch[i - 1]++;
  }
  leaf.insert(n);
  av.insert(1);
  while (dep > d && !leaf.empty() && !av.empty()) {
    while (!leaf.empty() && !av.empty()) {
      int it = *leaf.begin();
      leaf.erase(it);
      int node = -1, mn = -1e9;
      for (auto it2 : av) {
        if (lev[it2] < lev[it] && dep - (lev[it] - lev[it2] - 1) >= d &&
            (lev[it] - lev[it2] - 1) > 0) {
          if (mn < (lev[it] - lev[it2] - 1)) {
            mn = (lev[it] - lev[it2] - 1);
            node = it2;
          }
        }
      }
      if (node != -1) {
        dep -= (lev[it] - lev[node] - 1);
        ch[node]++;
        if (ch[node] > 1) av.erase(node);
        ch[par[it]]--;
        if (ch[par[it]] == 0) leaf.insert(par[it]);
        par[it] = node;
        lev[it] = lev[node] + 1;
      }
    }
  }
  if (dep != d) {
    cout << "NO\n";
  } else {
    cout << "YES\n";
    for (int i = 2; i <= n; i++) cout << par[i] << " ";
    cout << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}