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;
  vector<int> a(n + 1, 1);
  a[0] = 0;
  int sum_now = (n * (n - 1)) / 2 - d;
  for (int l = 2, r = a.size() - 1; l < r; r--) {
    while (l < r && (sum_now - r + l < 0 || a[l] == a[l - 1] * 2)) {
      l++;
    }
    if (l >= r) break;
    sum_now -= r;
    sum_now += l;
    a[l]++;
    a[r]--;
  }
  if (sum_now != 0) {
    cout << "NO\n";
    return;
  }
  cout << "YES\n";
  vector<int> pref_sums = a;
  for (int i = 1; i < pref_sums.size(); i++) {
    pref_sums[i] += pref_sums[i - 1];
  }
  for (int i = 2; i < pref_sums.size(); i++) {
    for (int k = 0; k < a[i]; k++) {
      cout << pref_sums[i - 2] + 1 + k / 2 << " ";
    }
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) solve();
  return 0;
}
