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 solve(int last, int rem) {
  if (last * 2 >= rem) return rem;
  return solve(last * 2, rem - last * 2) + rem;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n, d;
    cin >> n >> d;
    if (d < solve(1, n - 1) || d > (n * (n - 1)) / 2) {
      cout << "NO\n";
      continue;
    }
    vector<int> deg(n + 1, 1);
    deg[n] = 0;
    vector<int> par(n + 1), done(n + 1, 0), depth(n + 1, 0);
    for (int i = 1; i <= n; i++) par[i] = i - 1, depth[i] = i - 1;
    int sum = n * (n - 1) / 2;
    while (sum > d) {
      int leaf = -1;
      for (int i = 2; i <= n; i++) {
        if (done[i] || deg[i]) continue;
        leaf = i;
      }
      assert(leaf != -1);
      int parent = -1;
      for (int i = 1; i <= n; i++) {
        if (deg[i] == 2 || depth[i] > depth[leaf]) continue;
        if (depth[leaf] - depth[i] - 1 <= sum - d) {
          if (parent == -1 || depth[parent] > depth[i]) parent = i;
        }
      }
      assert(parent != -1);
      deg[parent]++;
      done[leaf] = 1;
      deg[par[leaf]]--;
      par[leaf] = parent;
      sum -= depth[leaf];
      depth[leaf] = depth[parent] + 1;
      sum += depth[leaf];
    }
    assert(sum == d);
    cout << "YES\n";
    for (int i = 2; i <= n; i++) cout << par[i] << " \n"[i == n];
  }
  return 0;
}