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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<image>
#include <bits/stdc++.h>
using namespace std;
void solve() {
  int n, d;
  cin >> n >> d;
  if (n * (n - 1) / 2 < d || n > 100 && d <= 3 * n) {
    cout << "NO\n";
    return;
  }
  int p[n + 1], dd[n + 1], cntd[n + 1];
  for (int i = 0; i < n; ++i) {
    dd[i] = i;
    p[i] = i - 1;
    cntd[i] = 1;
  }
  cntd[n - 1] = 0;
  int nd = n * (n - 1) / 2;
  bool u[n + 1];
  for (int i = 0; i < n; ++i) u[i] = false;
  while (nd != d) {
    bool ok = true, ok2 = false;
    for (int i = 0; i < n; ++i) ok &= u[i];
    if (ok) {
      cout << "NO\n";
      return;
    }
    for (int i = 0; i < n && !ok2; ++i) {
      if (u[i] || cntd[i]) continue;
      ok = false;
      for (int j = 0; j < n && !ok; ++j) {
        if (dd[j] == dd[i] - 2 && cntd[j] < 2) {
          cntd[p[i]]--;
          p[i] = j;
          cntd[j]++;
          dd[i]--;
          ok = true;
        }
      }
      if (!ok)
        u[i] = 1;
      else
        ok2 = true;
    }
    if (!ok2) {
      cout << "NO\n";
      return;
    }
    nd--;
  }
  cout << "YES\n";
  for (int i = 1; i < n; ++i) cout << p[i] + 1 << ' ';
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
}
