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;
#pragma GCC diagnostic ignored "-Wmissing-declarations"
int const maxn = 5005;
pair<int, int> bords[maxn];
int par[maxn];
void construct(int n, int d, const int shift = 1) {
  if (n == 1) return;
  d -= n - 1;
  for (int l = 1;; ++l) {
    int r = n - 1 - l;
    int lb = bords[l].first + bords[r].first;
    int rb = bords[l].second + bords[r].second;
    if (lb <= d && d <= rb) {
      int dl = bords[l].first;
      int coef = d - bords[r].second - dl;
      if (coef > 0) dl += coef;
      par[shift + 1] = shift;
      construct(l, dl, shift + 1);
      if (r > 0) {
        par[shift + l + 1] = shift;
        construct(r, d - dl, shift + l + 1);
      }
      break;
    }
  }
}
void solve() {
  int n, d;
  cin >> n >> d;
  if (d < bords[n].first || d > bords[n].second) {
    cout << "NO\n";
    return;
  }
  cout << "YES\n";
  fill(begin(par), end(par), 0);
  construct(n, d);
  for (int i = 2; i <= n; ++i) cout << par[i] << " \n"[i == n];
}
void precalc() {
  for (int n = 1; n < maxn; ++n) {
    int minval = 0;
    int cur = 0;
    for (int deep = 0; cur < n; ++deep) {
      int tmp = min(n - cur, 1 << deep);
      minval += deep * tmp;
      cur += tmp;
    }
    int maxval = n * (n - 1) / 2;
    bords[n] = {minval, maxval};
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  precalc();
  int t = 1;
  cin >> t;
  while (t--) solve();
}