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;
int n, d, m;
vector<int> levels, parent;
vector<vector<int> > on_level;
bool move_levels() {
  for (int i = n - 2; i >= 1; i--) {
    if ((levels[i] - 1) * 2 < levels[i + 1] + 1) continue;
    levels[i]--;
    levels[i + 1]++;
    return true;
  }
  return false;
}
void solve() {
  cin >> n >> d;
  int cur_depth = 0;
  int nn = n;
  levels.clear();
  for (int i = 0; nn > 0; i++) {
    levels.push_back(min(1 << i, nn));
    nn -= levels.back();
    cur_depth += levels.back() * i;
  }
  while ((int)levels.size() < n) levels.push_back(0);
  if (d < cur_depth) {
    cout << "NO\n";
    return;
  }
  while (cur_depth++ < d && move_levels())
    ;
  if (cur_depth <= d) {
    cout << "NO\n";
    return;
  }
  while (levels.back() == 0) levels.pop_back();
  m = levels.size();
  parent.assign(n, -1);
  on_level.assign(m, vector<int>());
  on_level[0].push_back(0);
  int cur = 1;
  for (int i = 1; i < m; i++) {
    int jj = 0;
    for (int j = 0; j < levels[i]; jj += j & 1, j++) {
      parent[cur] = on_level[i - 1][jj];
      on_level[i].push_back(cur++);
    }
  }
  cout << "YES\n";
  for (int i = 1; i < n; i++) cout << parent[i] + 1 << ' ';
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}