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 t, n, d;
int main() {
  cin >> t;
ed:
  while (t--) {
    cin >> n >> d;
    long long cnt[n], sum = 0;
    for (int i = 0; i < n; i++) cnt[i] = 1;
    sum = (n - 1) * n / 2;
    int last = n - 1, first = 1;
    if (d > sum) {
      cout << "NO" << '\n';
      goto ed;
    }
    while (1) {
      int x = d + last - sum;
      if (x >= first && cnt[x] < 2 * cnt[x - 1]) {
        cnt[x]++;
        cnt[last]--;
        break;
      } else {
        cnt[last]--;
        sum -= last;
        sum += first;
        cnt[first]++;
        if (!cnt[last]) last--;
        if (cnt[first] == 2 * cnt[first - 1]) first++;
        if (first >= last) {
          cout << "NO" << '\n';
          goto ed;
        }
      }
    }
    cout << "YES" << endl;
    bool bn;
    int j = 1;
    first = 1;
    for (int i = 1; i < n; i++) {
      bn = true;
      int q = 0;
      while (cnt[i] != q) {
        cout << j << " ";
        bn = !bn;
        if (bn) j++;
        q++;
      }
      first += cnt[i - 1];
      j = first;
    }
    cout << endl;
  }
  return 0;
}