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;
long long check(long long n) {
  long long i = 1, j = 0, k = 1;
  n--;
  for (; n != 0; k++) {
    if (i * 2 <= n) {
      n -= 2 * i;
      j += 2 * i * k;
    } else {
      j += n * k;
      n = 0;
    }
    i = i * 2;
  }
  return j;
}
void solve() {
  long long i, j, k, l, n, m, a, b, c, d;
  cin >> n >> d;
  if (check(n) > d || (n * (n - 1)) / 2 < d) {
    cout << "NO\n";
    return;
  }
  cout << "YES\n";
  long long h[n], p[n];
  c = (n * (n - 1)) / 2;
  for (i = 0; i < n; i++) {
    h[i] = 1;
    p[i] = -1;
  }
  a = 1;
  b = n - 1;
  while (c != d) {
    i = c - d;
    if (i >= (b - a)) {
      h[a]++;
      h[b]--;
      i = (b - a);
    } else {
      h[b - i]++;
      h[b]--;
    }
    c -= i;
    if (h[b] == 0) b--;
    if (h[a] == pow(2, a)) a++;
  }
  for (i = 1, j = 0, k = 1; i < n; k++) {
    for (l = 0; l < h[k]; l++, i++) {
      p[i] = j;
      if (l % 2 == 1) j++;
    }
    j = i - l;
  }
  for (i = 1; i < n; i++) cout << p[i] + 1 << " ";
  cout << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}