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>
#include <bits/stdc++.h>
using namespace std;
int maxn = 1e6;
int t, n, d;
int a, b;
vector<int> arr(maxn + 1);
vector<int> fenwick(maxn + 1);
int getsum(int idx) {
  int s = 0;
  while (idx > 0) {
    s += fenwick[idx];
    idx -= idx & -idx;
  }
  return s;
}
void update(int idx, int delta) {
  while (idx <= n) {
    fenwick[idx] += delta;
    idx += idx & -idx;
  }
}
void solve() { return; }
int ipow(int base, int exp) {
  if (exp == 0) return 1;
  if (exp == 1) return base;
  int temp = ipow(base, exp / 2);
  if (exp % 2 == 0)
    return temp * temp;
  else
    return base * temp * temp;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  vector<int> a(5001);
  int cur = 0;
  int cnt = 0;
  for (int i = 1; i <= 5000; i++) {
    a[i] = cur;
    cnt++;
    if (cnt == ipow(2, cur)) {
      cnt = 0;
      cur++;
    }
  }
  vector<int> b(5001);
  for (int i = 1; i <= 5000; i++) {
    b[i] = b[i - 1] + a[i];
  }
  cin >> t;
  while (t--) {
    cin >> n >> d;
    if (d < b[n] || d > n * (n - 1) / 2) {
      cout << "NO"
           << "\n";
    } else {
      cout << "YES"
           << "\n";
      vector<vector<int>> tree(n);
      for (int i = 0; i < n; i++) {
        tree[i].push_back(i + 1);
      }
      int total = n * (n - 1) / 2;
      int cur_num = n;
      int cur_dep = cur_num - 1;
      while (total > d) {
        if (cur_dep - 1 > 20 ||
            tree[cur_dep - 1].size() < ipow(2, cur_dep - 1)) {
          tree[cur_dep].pop_back();
          tree[cur_dep - 1].push_back(cur_num);
          cur_dep--;
          total--;
        } else {
          cur_num--;
          cur_dep = cur_num - 1;
        }
      }
      vector<int> parents(n + 1);
      for (int i = n - 1; i >= 1; i--) {
        for (int j = 0; j < tree[i].size(); j++) {
          parents[tree[i][j]] = tree[i - 1][j / 2];
        }
      }
      for (int i = 2; i <= n; i++) {
        cout << parents[i] << " ";
      }
      cout << "\n";
    }
  }
  return 0;
}
