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;
const long long int inf = 1e18, M = 1e9 + 7;
const long long int N = 5e3 + 10, LN = 16;
vector<long long int> v(N, 0), val(LN);
void compute() {
  val[0] = 1;
  for (long long int i = 1; i < LN; ++i) val[i] = 2ll * val[i - 1];
}
long long int func(long long int n) {
  long long int tmp = n;
  long long int res = 0;
  for (long long int i = 0;; ++i) {
    if (val[i] >= tmp) {
      res += tmp * i;
      break;
    }
    res += i * val[i];
    tmp -= val[i];
  }
  return res;
}
void solve() {
  long long int n, d;
  cin >> n >> d;
  long long int sum = (n * (n - 1)) / 2;
  if (d < func(n) || sum < d) {
    cout << "NO";
    return;
  }
  cout << "YES\n";
  for (long long int i = 0; i < n; ++i) v[i] = 1;
  long long int cur = 1;
  for (long long int i = n - 1; i >= 0; --i) {
    --v[i];
    if (sum - i + cur > d) {
      sum += -i + cur;
      ++v[cur];
      if (v[cur] == val[cur]) ++cur;
    } else {
      long long int tmp = d + i - sum;
      ++v[tmp];
      break;
    }
  }
  long long int cnt;
  sum = 0;
  for (long long int i = 1; i < n && v[i]; ++i) {
    cnt = sum + 1;
    for (long long int j = 0; j < v[i]; ++cnt) {
      if (j == v[i] - 1)
        cout << cnt << " ", ++j;
      else
        cout << cnt << " " << cnt << " ", j += 2;
    }
    sum += v[i - 1];
  }
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  compute();
  long long int t = 1;
  cin >> t;
  while (t--) {
    solve();
    cout << "\n";
  }
  return 0;
}