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 M = 5e4 + 4;
int ans[M];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n, d;
    cin >> n >> d;
    int now = (1 + n - 1) * (n - 1) / 2;
    if (d > now) {
      cout << "NO" << endl;
    } else {
      for (int i = 2; i <= n; i++) ans[i] = i - 1;
      int flag = 1;
      queue<int> que, que2;
      while (!que.empty()) que.pop();
      while (!que2.empty()) que2.pop();
      que.push(1);
      int head = 1, tail = n;
      while (now != d) {
        if (head + 1 == tail) {
          flag = 0;
          break;
        }
        int dec = tail - head - 1;
        if (dec > now - d) {
          ans[tail] = tail - (now - d) - 1;
          break;
        }
        if (que.empty()) {
          while (!que2.empty()) que.push(que2.front()), que2.pop();
          head++;
          que.push(head);
          continue;
        }
        ans[tail] = que.front();
        que.pop();
        que2.push(tail);
        que2.push(tail);
        tail--;
        now -= dec;
      }
      if (flag) {
        cout << "YES" << endl;
        for (int i = 2; i <= n; i++) cout << ans[i] << ' ';
        cout << '\n';
      } else {
        cout << "NO" << endl;
      }
    }
  }
}