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:

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#include <bits/stdc++.h>
template <class C>
inline void log_c(const C& c) {}
template <class C>
inline int sz(const C& c) {
  return static_cast<int>(c.size());
}
using namespace std;
using pii = pair<int, int>;
using num = int64_t;
using pll = pair<num, num>;
const std::string eol = "\n";
inline int level_capacity(int d) { return 1 << d; }
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  vector<int> v;
  for (int q_idx = 0; q_idx < (t); ++q_idx) {
    int n, d;
    cin >> n >> d;
    v.resize(n);
    fill(begin(v), end(v), 1);
    int l = 1;
    int r = n - 1;
    int s = n * (n - 1) / 2;
    if (s > d) {
      for (; l < r; --r) {
        const int delta = min(s - d, r - l);
        l = r - delta;
        --v[r];
        ++v[l];
        if (v[l] >= level_capacity(l)) ++l;
        s -= delta;
      }
    }
    cout << (s == d ? "YES" : "NO") << eol;
    if (s == d) {
      int s = 2;
      for (int k = 1; k < n && v[k]; ++k) {
        for (int j = 0; j < (v[k]); ++j) {
          cout << (s - v[k - 1] + j / 2) << ' ';
        }
        s += v[k];
      }
      cout << eol;
    }
  }
  return 0;
}
