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() {
  int q;
  cin >> q;
  while (q--) {
    int n, d;
    cin >> n >> d;
    int low = 0;
    for (int i = 1; i <= n; i++) low += (int)log2(i);
    if (low > d) {
      cout << "NO" << endl;
      continue;
    }
    vector<int> dep(n), par(n, -1), child(n, 1);
    child.back() = 0;
    vector<bool> bad(n, 0);
    iota(dep.begin(), dep.end(), 0);
    for (int i = 0; i < n - 1; i++) par[i + 1] = i;
    int tmp = (n - 1) * n / 2;
    while (tmp > d) {
      int del = -1, deldep = 1 << 30;
      for (int i = 0; i < n; i++) {
        if (bad[i] || child[i]) continue;
        if (deldep > dep[i]) del = i, deldep = dep[i];
      }
      if (del == -1) break;
      int newp = -1;
      for (int i = 0; i < n; i++) {
        if (child[i] < 2 && dep[i] + 2 == dep[del]) {
          newp = i;
          break;
        }
      }
      if (newp == -1) {
        bad[del] = true;
        continue;
      }
      child[par[del]]--;
      par[del] = newp;
      child[newp]++;
      dep[del]--;
      tmp--;
    }
    if (tmp != d) {
      cout << "NO" << endl;
      continue;
    }
    cout << "YES" << endl;
    for (int i = 1; i < n; i++) cout << par[i] + 1 << " \n"[i == n - 1];
  }
  return 0;
}