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 Down_sum(vector<int> &p, vector<int> &l, vector<int> &d, vector<int> &b,
              int &sum, int n) {
  bool f = true, f2 = true;
  while (f) {
    f2 = true;
    f = true;
    int second = 0, md = 1e9;
    for (int i = 2; i <= n; ++i)
      if (b[i] != 1 && l[i] == 1 && d[i] < md) {
        second = i;
        md = d[i];
        f2 = false;
      }
    for (int i = 1; i <= n; ++i)
      if (l[i] != -1 && d[i] == d[second] - 2) {
        --sum;
        --l[i];
        --d[second];
        ++l[p[second]];
        p[second] = i;
        f = false;
        break;
      }
    if (f) b[second] = 1;
    if (f2) {
      sum = -1;
      break;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  int t = 100;
  cin >> t;
  while (t--) {
    int n, D;
    cin >> n >> D;
    vector<int> p(n + 1, 0);
    vector<int> d(n + 1, 0), s2(n + 1, 0);
    vector<int> l(n + 1, 1);
    vector<int> b(n + 1, 0);
    int sum = 0, minD = 0;
    for (int i = 2; i <= n; ++i) {
      s2[i] = s2[i / 2] + 1;
      minD += s2[i];
      --l[i - 1];
      p[i] = i - 1;
      d[i] = d[i - 1] + 1;
      sum += d[i];
    }
    if (sum < D || minD > D) {
      cout << "NO\n";
      continue;
    }
    while (sum > D) {
      Down_sum(p, l, d, b, sum, n);
    }
    if (sum == -1) {
      cout << "NO\n";
      continue;
    }
    cout << "YES\n";
    for (int i = 2; i <= n; ++i) cout << p[i] << " ";
    cout << "\n";
  }
  return 0;
}