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 N = 5005;
int low[N];
int high[N];
void precalc() {
  int cur;
  int tk;
  for (int i = 1; i < N; i++) {
    cur = i;
    high[i] = (i * (i - 1)) / 2;
    for (int j = 0; j < 15; j++) {
      tk = min((1 << j), cur);
      low[i] += tk * j;
      cur -= tk;
    }
  }
}
bool is_good(int n, int d) { return (d >= low[n] && d <= high[n]); }
vector<int> solve(int n, int d) {
  if (n == 1) return {};
  if (is_good(n - 1, d - n + 1)) {
    vector<int> yy = solve(n - 1, d - n + 1);
    vector<int> ret;
    ret.push_back(1);
    for (auto x : yy) ret.push_back(x + 1);
    return ret;
  }
  int r, f;
  for (int l = 1; l + 1 < n; l++) {
    for (int k = low[l]; k <= high[l]; k++) {
      r = n - l - 1;
      f = d - r - k - l;
      if (is_good(r, f)) {
        vector<int> lf = solve(l, k);
        vector<int> rf = solve(r, f);
        vector<int> res;
        res.push_back(1);
        for (auto x : lf) res.push_back(x + 1);
        res.push_back(1);
        for (auto x : rf) res.push_back(x + 2 + lf.size());
        return res;
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  precalc();
  int tc;
  cin >> tc;
  for (int t = 0; t < tc; t++) {
    int n, d;
    cin >> n >> d;
    if (!is_good(n, d)) {
      cout << "NO\n";
    } else {
      cout << "YES\n";
      vector<int> res = solve(n, d);
      for (auto x : res) cout << x << " ";
      cout << "\n";
    }
  }
  return 0;
}