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;
using ll = long long;
using ld = long double;
using u64 = uint64_t;
using u32 = uint32_t;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;
using vvi = vector<vi>;
using vpi = vector<pi>;
static constexpr int INF = (int)1e9 + 5;
static constexpr ll INFL = (ll)1e18 + 5;
static mt19937 rng((u32)chrono::duration_cast<chrono::nanoseconds>(
                       chrono::high_resolution_clock::now().time_since_epoch())
                       .count());
struct Layer {
  vi nodes;
  int open = 0;
  int deg_open = 1;
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    int n, d;
    cin >> n >> d;
    vi p(n - 1, -1);
    vector<Layer> l(n);
    for (int i = 0; i < (n); ++i) l[i].nodes.push_back(i);
    for (int i = 0; i < (n - 1); ++i) p[i] = i;
    d -= n * (n - 1) / 2;
    for (int i = n - 1; i > 0; --i) {
      if (!d || ((int)(l[i].nodes).size()) != 1) break;
      int par = max(0, i + d - 1);
      while (par < i - 1 && l[par].open == ((int)(l[par].nodes).size())) ++par;
      if (par >= i - 1) break;
      int v = l[i].nodes[0];
      l[i].nodes.pop_back();
      int pv = l[par].nodes[l[par].open];
      p[v - 1] = pv;
      l[par + 1].nodes.push_back(v);
      d += i - par - 1;
      if (++l[par].deg_open == 2) {
        l[par].deg_open = 0;
        ++l[par].open;
      }
    }
    if (d)
      cout << "NO\n";
    else {
      cout << "YES\n";
      for (auto& i : p) cout << (i + 1) << ' ';
      cout << '\n';
    }
  }
  return 0;
}