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;
const int N = (int)5e3 + 11;
int n, d;
void solve() {
  cin >> n >> d;
  int mx = n * (n - 1) / 2;
  int mn = 0, cnt = 0, deg = 0;
  while (cnt < n) {
    int cur = (1 << deg);
    cur = min(cur, n - cnt);
    cnt += cur;
    mn += deg * cur;
    deg++;
  }
  if (mn <= d and d <= mx) {
    cout << "YES\n";
    int cnt[n + 11];
    memset(cnt, 0, sizeof(cnt));
    for (int i = 0; i < n; i++) cnt[i] = 1;
    int de = mx - d;
    while (de > 0) {
      for (int i = n - 1; i >= 1; i--) {
        int x = i - 2;
        if (x < 0 or cnt[i + 1] > (cnt[i] - 1) * 2 or
            cnt[x + 1] + 1 > cnt[x] * 2)
          continue;
        cnt[i]--;
        cnt[x + 1]++;
        de--;
        if (de == 0) break;
      }
    }
    vector<int> pr;
    pr.push_back(1);
    pr.push_back(1);
    int cur = 2;
    for (int i = 1; i < n; i++) {
      vector<int> g;
      while (cnt[i] > 0) {
        cout << pr.back() << " ";
        g.push_back(cur);
        g.push_back(cur);
        pr.pop_back();
        cnt[i]--;
        cur++;
      }
      pr = g;
    }
    cout << endl;
  } else
    cout << "NO\n";
}
int main() {
  ;
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int tt = 1;
  cin >> tt;
  for (int i = 1; i <= tt; i++) {
    solve();
  }
  return 0;
}