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 xd[4] = {0, 1, 0, -1}, yd[4] = {1, 0, -1, 0};
template <typename t>
bool ckmin(t& a, const t& b) {
  return a > b ? a = b, true : false;
}
template <typename t>
bool ckmax(t& a, const t& b) {
  return a < b ? a = b, true : false;
}
int sums[5010];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  sums[0] = 0;
  for (auto((i)) = (0); ((i)) < ((5000)); ++((i)))
    sums[i + 1] = sums[i] + i + 1;
  int tcs;
  cin >> tcs;
  while (tcs--) {
    int n, d;
    cin >> n >> d;
    int maximum = sums[n - 1];
    int minimum = 0;
    for (int nodes = n, depth = 0; nodes;
         nodes = max(0, nodes - (1 << depth++)))
      minimum += min(nodes, 1 << depth) * depth;
    if (minimum > d || d > maximum) {
      cout << "NO" << endl;
      continue;
    }
    vector<int> parent(n, -1), childs(n, 1);
    childs.back() = 0;
    set<pair<int, int> > goodLeaves;
    goodLeaves.emplace(n - 1, n - 1);
    vector<int> depth(n);
    iota((depth).begin(), (depth).end(), 0);
    iota((parent).begin(), (parent).end(), -1);
    int val = sums[n - 1];
    while (val > d) {
      auto [de, leave] = *goodLeaves.begin();
      goodLeaves.erase(goodLeaves.begin());
      int p = -1;
      if (depth[leave] > 1)
        for (auto((i)) = (0); ((i)) < ((n)); ++((i)))
          if (depth[i] == de - 2 && childs[i] < 2) p = i;
      if (p == -1) continue;
      if (--childs[parent[leave]] == 0)
        goodLeaves.emplace(depth[parent[leave]], parent[leave]);
      ++childs[p];
      parent[leave] = p;
      --depth[leave];
      val--;
      goodLeaves.emplace(de - 1, leave);
    }
    cout << "YES" << endl;
    for (auto(i) = (1); (i) < (n); ++(i)) cout << parent[i] + 1 << ' ';
    cout << endl;
  }
  return 0;
}