Problem: 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>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long T;
  cin >> T;
  while (T--) {
    long long n, d;
    cin >> n >> d;
    vector<long long> par(n + 1, 0);
    vector<long long> child(n + 1, 0);
    vector<vector<long long>> levels;
    levels.resize(n + 1);
    long long sum = 0;
    long long mx = n - 1;
    long long currlevel = 0;
    for (long long i = 1; i < n; i++) {
      child[i] = i + 1;
      par[i + 1] = i;
      levels[i - 1].push_back(i);
      sum += (i);
    }
    levels[n - 1].push_back(n);
    while (sum >= d) {
      while (pow(2, currlevel) == levels[currlevel].size()) {
        currlevel++;
      }
      if (mx <= currlevel) {
        break;
      } else {
        long long diff = mx - currlevel;
        if (diff <= sum - d) {
          long long curr = levels[mx][0];
          levels[currlevel].push_back(curr);
          levels[mx].erase(levels[mx].begin());
          sum -= diff;
        } else {
          long long curr = levels[mx][0];
          levels[mx - sum + d].push_back(curr);
          levels[mx].erase(levels[mx].begin());
          sum = d;
        }
        mx--;
      }
    }
    if (sum != d) {
      cout << "NO" << endl;
    } else {
      cout << "YES" << endl;
      long long cnt = 1;
      vector<vector<long long>> Tree;
      Tree.resize(n + 1);
      for (long long i = 0; i < n + 1; i++) {
        for (long long j = 0; j < levels[i].size(); j++) {
          Tree[i].push_back(cnt);
          cnt++;
        }
      }
      for (long long i = 1; i < n + 1; i++) {
        for (long long j = 0; j < Tree[i].size(); j++) {
          cout << Tree[i - 1][j / 2] << " ";
        }
      }
      cout << endl;
    }
  }
  return 0;
}