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 layer[5005], num[5005];
void solve() {
  memset(num, 0, sizeof num);
  int n, d;
  cin >> n >> d;
  int sum = n * (n - 1) / 2, dep = 0, minn = 0;
  num[0] = 1;
  for (int i = 2; i <= n; i++) {
    if ((i & (i - 1)) == 0) dep++;
    minn += dep;
    layer[i - 1] = i - 1;
    num[i - 1] = 1;
  }
  if (d < minn || d > sum) {
    cout << "NO" << endl;
    return;
  }
  cout << "YES" << endl;
  dep = 1;
  for (int i = n - 1; i > 0 && sum > d; i--) {
    sum -= i;
    num[i]--;
    if (sum > d) {
      layer[i] = dep;
      sum += dep;
      if (++num[dep] == (1 << dep)) dep++;
    } else {
      int dif = d - sum;
      if (dif >= dep) {
        layer[i] = dif;
        sum += dif;
        num[dif]++;
        break;
      } else {
        layer[i] = dep;
        sum += dep;
        if (++num[dep] == (1 << dep)) dep++;
      }
    }
  }
  int id(2), fid(1);
  for (int i = 1; num[i]; i++) {
    for (int j = 0; j < num[i]; j++) {
      layer[id++] = fid + j / 2;
    }
    fid += num[i - 1];
  }
  for (int i = 2; i <= n; i++) {
    cout << layer[i] << " ";
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}