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;
  int D = d;
  vector<int> p(n);
  vector<int> h(n);
  for (int i = 1; i < n; ++i) {
    p[i] = (i + 1) / 2 - 1;
    h[i] = h[(i + 1) / 2 - 1] + 1;
    d -= h[i];
  }
  if (d < 0) {
    cout << "NO";
    return;
  }
  while (d) {
    vector<int> cnt(n);
    for (int i = 0; i < n; ++i) {
      ++cnt[p[i]];
    }
    vector<vector<int>> fr(D + 1);
    for (int i = 0; i < n; ++i) {
      if (cnt[i] < 2) {
        fr[h[i]].push_back(i);
      }
    }
    bool ok = false;
    for (int i = 0; i < n; ++i) {
      if (cnt[i] == 0 and fr[h[i]].size() > 1) {
        for (int it : fr[h[i]]) {
          if (it != i) {
            p[i] = it;
            ++h[i];
            --d;
            ok = true;
            break;
          }
        }
      }
      if (ok) {
        break;
      }
    }
    if (!ok) {
      cout << "NO";
      return;
    }
  }
  cout << "YES\n";
  for (int i = 1; i < n; ++i) {
    cout << p[i] + 1 << ' ';
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
    cout << '\n';
  }
  return 0;
}
