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>
#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, d;
    cin >> n >> d;
    vector<int> lev(n + 1, 1);
    int l = 2, r = n;
    int p = (n * (n - 1) / 2) - d;
    while (l < r) {
      while (l < r && (p - (r - l) < 0 || lev[l] == lev[l - 1] * 2)) l++;
      if (l < r) {
        p -= (r - l);
        lev[l]++;
        r--;
      }
    }
    if (p != 0)
      cout << "NO";
    else {
      cout << "YES" << endl;
      int pre = 1;
      for (int i = 2; i <= l; i++) {
        for (int j = 0; j < lev[i]; j++) {
          cout << pre + j / 2 << " ";
        }
        pre += lev[i - 1];
      }
    }
    cout << endl;
  }
  return 0;
}
