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;
void solve() {
  long long n, d;
  cin >> n >> d;
  long long up = (n * (n - 1)) / 2;
  long long down = 0, mm = 0;
  long long lim[n + 1], par[n + 1], dep[n + 1];
  for (long long i = 1; i <= n; i++) {
    if (!(i & (i - 1))) {
      mm++;
    }
    down += mm - 1;
    lim[mm - 1]++;
  }
  for (long long i = 0; i < n; i++) {
    par[i] = i - 1;
    dep[i] = 1;
  }
  if (d < down || d > up) {
    cout << "NO\n";
    return;
  }
  cout << "YES\n";
  for (long long i = n - 1; i > 0; i--) {
    for (long long j = 1; j < i; j++) {
      if (up - i + j >= d && dep[j] < pow(2, j)) {
        up = up - i + j;
        dep[j]++;
        dep[i]--;
        break;
      }
    }
  }
  queue<pair<long long, long long> > q;
  q.push({0, 0});
  mm = 1;
  for (long long i = 1; i < n; i++) {
    if (dep[i] == 0) break;
    while (dep[i]--) {
      long long t = q.front().first, lev = q.front().second;
      q.pop();
      par[mm] = t;
      q.push({mm, lev + 1});
      mm++;
      if (dep[i] > 0) {
        par[mm] = t;
        q.push({mm, lev + 1});
        mm++;
        dep[i]--;
      }
      if (dep[i] == 0) {
        while (q.front().second == lev) {
          q.pop();
        }
      }
    }
  }
  for (long long i = 1; i < n; i++) {
    cout << par[i] + 1 << " ";
  }
  cout << endl;
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
}