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>
#include <bits/stdc++.h>
using namespace std;
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, d;
    cin >> n >> d;
    long long dmax = n * (n - 1) / 2;
    long long dmin[n];
    long long temp = 1;
    long long index = 1;
    long long val = 1;
    while (index < n) {
      for (long long j = 1; j <= 2 * temp && index < n; j++) {
        dmin[index] = val;
        index++;
      }
      temp *= 2;
      val += 1;
    }
    long long dmini = 0;
    for (long long i = 1; i < n; i++) dmini += dmin[i];
    if (d < dmini || d > dmax)
      cout << "NO\n";
    else {
      cout << "YES\n";
      long long orh[n];
      for (int i = 0; i < n; i++) {
        orh[i] = i;
      }
      long long minh[n];
      long long i = 1;
      index = n - 1;
      while (index >= 0 && i < n) {
        for (long long j = 1; j <= pow(2, i) - 1 && i < n && index >= 0; j++) {
          minh[index] = min(index, i);
          index--;
        }
        i++;
      }
      long long jhc = dmax - d;
      for (long long i = n - 1; i >= 0 && jhc > 0; i--) {
        long long x = orh[i] - minh[i];
        for (long long j = 1; j <= x && jhc > 0; j++) {
          jhc--;
          orh[i]--;
        }
      }
      long long maxhi = INT_MIN;
      for (long long i = 0; i < n; i++) {
        if (orh[i] > maxhi) maxhi = orh[i];
      }
      vector<long long> v[maxhi + 1];
      for (long long i = 0; i < n; i++) {
        v[orh[i]].push_back(i);
      }
      long long p[n];
      for (long long i = 1; i < maxhi + 1; i++) {
        for (long long j = 0; j < v[i].size(); j++) {
          long long num = v[i][j];
          long long upari = j / 2;
          p[num] = v[i - 1][upari] + 1;
        }
      }
      for (long long i = 1; i < n; i++) cout << p[i] << " ";
      cout << endl;
    }
  }
  return 0;
}
