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;
long long int c, n, d, p[5050], cnt[5050];
stack<long long int> s[5050];
int main() {
  ios::sync_with_stdio(false);
  cin >> c;
  while (c--) {
    long long int S = 0, last;
    for (int i = 1; i <= n; i++)
      while (!s[i].empty()) s[i].pop();
    cin >> n >> d;
    last = n;
    cnt[0] = 1;
    for (int i = 1; i <= n; i++) p[i] = i - 1, S += i - 1, s[i].push(i);
    for (int i = 1; i <= n; i++) cnt[i]--;
    {
      while (S > d) {
        bool ok = false;
        for (int i = 1; i < last; i++) {
          if (s[i].size() > 0 and S - ((last - 1) - i) >= d) {
            p[last] = s[i].top(), s[i].pop(), S -= ((last - 1) - i);
            s[i + 1].push(last), s[i + 1].push(last);
            last--;
            ok = true;
            break;
          }
        }
        if (!ok) break;
      }
      if (S == d) {
        cout << "YES\n";
        for (int i = 2; i <= n; i++) cout << p[i] << " ";
        cout << endl;
      } else
        cout << "NO\n";
    }
  }
}
