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>
#pragma GCC optimize("Ofast")
using namespace std;
int mi[5005], mx[5005];
bool check(int p, int x) { return (mi[p] <= x && mx[p] >= x); }
vector<int> a[5005];
int ans[5005];
void solve() {
  int n, d;
  cin >> n >> d;
  if (!check(n, d)) {
    cout << "NO\n";
    return;
  }
  for (int i = 1; i <= n; i++) a[i].clear(), a[i].push_back(i);
  int len = n, t = mx[n], now, s;
  t -= d;
  while (t) {
    while (!a[len].size()) len--;
    s = a[len].back();
    a[len].pop_back();
    now = len;
    while (now > 1 && t) {
      t--;
      now--;
    }
    while (a[now].size() >= a[now - 1].size() * 2) {
      now++;
      t++;
    }
    a[now].push_back(s);
  }
  for (int i = 2; i <= len; i++) {
    for (int j = 0; j < a[i].size(); j++) {
      ans[a[i][j]] = a[i - 1][j / 2];
    }
  }
  cout << "YES\n";
  for (int i = 2; i <= n; i++) cout << ans[i] << ' ';
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int cnt = 0, p = 0, sum = 0;
  for (int i = 1; i <= 5000; i++) {
    sum += p;
    cnt++;
    mi[i] = sum;
    if (cnt == (1 << p)) {
      p++;
      cnt = 0;
    }
  }
  for (int i = 1; i <= 5000; i++) mx[i] = i * (i - 1) / 2;
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}