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;
void sep() { cout << "-------------BLOCK-----------------\n"; }
template <typename T>
void debugv(vector<T> arr) {
  for (int i = 0; i < arr.size() - 1; i++) cout << arr[i] << "p" << i << " ";
  cout << "\n";
}
template <typename T>
void debugarr(T arr, int n) {
  for (int i = 0; i < n; i++) cout << arr[i] << " ";
  cout << "\n";
}
long long __min(long long a, long long b) { return a < b ? a : b; }
long long __max(long long a, long long b) { return a > b ? a : b; }
const int maxn = 4e6 + 100;
long long n, m, t, k, q;
unordered_map<long long, long long> u;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> t;
  while (t--) {
    cin >> n >> k;
    long long ub = (n * (n - 1)) / 2;
    long long lb = 0;
    long long ac = 0;
    for (int i = 1; i <= n; i++) {
      if (!(i & (i - 1))) ac++;
      lb += ac - 1;
    }
    if (k < lb || k > ub) {
      cout << "NO\n";
      continue;
    }
    vector<long long> par(n + 1, -1);
    iota(par.begin(), par.end(), -1);
    vector<long long> child(n + 1, 1);
    vector<long long> dep(n + 1, 0);
    iota(dep.begin(), dep.end(), 0);
    vector<bool> bad(n + 1, false);
    child[n - 1] = 0;
    while (ub > k) {
      long long v = -1;
      for (int i = 0; i < n; i++) {
        if (!bad[i] && child[i] == 0 && (v == -1 || dep[i] < dep[v])) {
          v = i;
        }
      }
      long long p = -1;
      for (int i = 0; i < n; i++) {
        if (child[i] <= 1 && dep[i] < dep[v] - 1 &&
            (p == -1 || dep[p] < dep[i])) {
          p = i;
        }
      }
      if (p == -1) {
        bad[v] = true;
        continue;
      }
      assert(dep[v] - dep[p] == 2);
      --child[par[v]];
      dep[v]--;
      ++child[p];
      par[v] = p;
      ub--;
    }
    cout << "YES\n";
    for (int i = 1; i < n; i++) cout << par[i] + 1 << " ";
    cout << endl;
  }
}