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>
#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template <typename T>
bool MinPlace(T &a, const T &b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
bool MaxPlace(T &a, const T &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <typename S, typename T>
ostream &operator<<(ostream &out, const pair<S, T> &p) {
  out << "{" << p.first << ", " << p.second << "}";
  return out;
}
template <typename T>
ostream &operator<<(ostream &out, const vector<T> &v) {
  out << "[";
  for (int i = 0; i < (int)v.size(); i++) {
    out << v[i];
    if (i != (int)v.size() - 1) out << ", ";
  }
  out << "]";
  return out;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t, n = 5000, d;
  vector<int> minm(n + 1, 0);
  for (int i = 1; i <= n; i++) minm[i] = minm[i - 1] + (31 - __builtin_clz(i));
  cin >> t;
  while (t--) {
    cin >> n >> d;
    if (d < minm[n] || (d > n * (n - 1) / 2)) {
      cout << "NO\n";
      continue;
    }
    d = n * (n - 1) / 2 - d;
    vector<vector<pair<int, int>>> byLevel(n);
    vector<int> lvl(n + 1, 0), pa(n + 1, 0);
    for (int i = 0; i < n; i++) {
      byLevel[i].push_back(make_pair(i + 1, 1));
      lvl[i + 1] = i;
      pa[i + 1] = i;
    }
    byLevel[n - 1][0].second++;
    for (int i = n, up = 0; d and i > 2; i--) {
      d -= lvl[i] - (up + 1);
      if (d < 0) {
        up -= d;
        d = 0;
      }
      lvl[i] = up + 1;
      byLevel[up + 1].push_back({i, 2});
      pa[i] = byLevel[up].back().first;
      if (!(--byLevel[up].back().second)) {
        byLevel[up].pop_back();
        up += !(byLevel[up].size());
      }
    }
    cout << "YES\n";
    for (int i = 2; i <= n; i++) cout << pa[i] << " ";
    cout << "\n";
  }
};
