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;
const int N = 5e3 + 10;
int l[N], r[N], p[N], d[N];
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n, z;
    cin >> n >> z;
    set<pair<int, int>> s;
    int sum = 0;
    for (int i = 1; i < n; ++i) {
      sum += log2(i);
      r[i] = 0;
      p[i] = d[i] = i - 1;
      l[i] = i + 1;
      s.insert({i - 1, i});
    }
    if (sum > z) {
      cout << "NO\n";
      continue;
    }
    set<int> leaves;
    leaves.insert(n);
    p[n] = d[n] = n - 1;
    int k = n * (n - 1);
    k >>= 1;
    k -= z;
    if (k < 0) {
      cout << "NO\n";
      continue;
    }
    bool b = 0;
    while (k) {
      b = 1;
      for (auto &i : leaves) {
        auto it = s.lower_bound({d[i] - 2, -1});
        if (it != s.end() && it->first == d[i] - 2 && it->second != i) {
          if (d[it->second] != it->first) {
            s.erase(it);
            continue;
          }
          b = 0;
          int x = it->second;
          if (!l[x])
            l[x] = i;
          else
            r[x] = i;
          if (leaves.find(x) != leaves.end()) leaves.erase(leaves.find(x));
          --d[i];
          --k;
          if (r[x] && l[x]) s.erase(it);
          if (l[p[i]] == i)
            l[p[i]] = 0;
          else
            r[p[i]] = 0;
          if (!l[p[i]] && !r[p[i]])
            leaves.insert(p[i]);
          else
            s.insert({d[p[i]], p[i]});
          p[i] = x;
          s.insert({d[i], i});
          break;
        } else {
          b = 0;
          leaves.erase(leaves.find(i));
          break;
        }
      }
      if (b) break;
    }
    if (b)
      cout << "NO";
    else {
      cout << "YES\n";
      for (int i = 2; i <= n; ++i) cout << p[i] << ' ';
    }
    cout << '\n';
  }
  return 0;
}
