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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n, d;
    cin >> n >> d;
    long long kd[n + 5];
    long long hlt = (n * (n - 1)) / 2;
    long long llt = 0;
    long long depth[n + 5];
    memset(depth, 0, sizeof(depth));
    kd[1] = 0;
    depth[0]++;
    for (long long i = 2; i <= n; i++) {
      kd[i] = kd[i / 2] + 1;
      llt += kd[i];
      depth[kd[i]]++;
    }
    if (d < llt or d > hlt) {
      cout << "NO" << '\n';
      continue;
    }
    long long diff = d - llt;
    long long h = n - 1;
    while (depth[h] == 0) {
      h--;
    }
    long long pm = h++;
    while (diff > 0) {
      long long pmi;
      while (depth[pm] == 1) {
        pm--;
      }
      pmi = pm;
      while (pmi < h && diff > 0) {
        depth[pmi]--;
        depth[pmi + 1]++;
        pmi++;
        diff--;
      }
      if (pmi == h) {
        h++;
      }
    }
    cout << "YES" << '\n';
    long long x = 0, y = 1;
    for (long long i = 1; i < n; i++) {
      for (long long j = 0; j < depth[i]; j++) {
        cout << (x + j / 2 + 1) << " ";
      }
      x = y;
      y += depth[i];
    }
    cout << '\n';
  }
  return 0;
}