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;
int const N = 5007;
int level[N];
int p[N];
queue<pair<int, int> > q;
int getMinimum(int n) {
  n--;
  int res = 0;
  int ctr = 1;
  for (int i = 2; n != 0; i *= 2, ctr++) res += min(i, n) * ctr, n -= min(i, n);
  return res;
}
int getMaximum(int n) { return (n * (n - 1)) / 2; }
int main() {
  int t;
  cin >> t;
  int n, d;
  for (int i = 0; i < t; ++i) {
    cin >> n >> d;
    if (d < getMinimum(n) || getMaximum(n) < d) {
      cout << "NO\n";
      continue;
    }
    cout << "YES\n";
    for (int i = 0; i < n; ++i) level[i] = 1;
    int dif = getMaximum(n) - d;
    int idx = 1;
    for (int i = n - 1; i >= 0; --i) {
      if (i - idx < dif)
        level[i]--, level[idx]++, dif -= (i - idx);
      else
        level[i]--, level[i - dif]++, dif = 0;
      if (level[idx] == 2 * level[idx - 1]) idx++;
      if (dif == 0) break;
    }
    while (q.size()) q.pop();
    q.push(pair<int, int>(1, 1));
    int node = 2;
    for (int i = 1; i < n;) {
      if (level[i] == 0) {
        ++i;
        continue;
      }
      while (i > q.front().first) q.pop();
      if (level[i] == 1)
        p[node] = q.front().second, q.push(pair<int, int>(i + 1, node)), node++,
        q.pop(), level[i] = 0;
      else {
        p[node] = q.front().second, q.push(pair<int, int>(i + 1, node)), node++;
        p[node] = q.front().second, q.push(pair<int, int>(i + 1, node)), node++;
        q.pop();
        level[i] -= 2;
      }
    }
    for (int i = 2; i < n; ++i) cout << p[i] << ' ';
    cout << p[n] << '\n';
  }
  return 0;
}