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 a[110];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 0, n = 0, d = 0, c = 0, x = 0, flag = 0, f = 0;
  cin >> t;
  while (t--) {
    memset(a, 0, sizeof(a));
    cin >> n >> d;
    a[0] = 1;
    c = 1;
    for (int i = 1; i <= 10; ++i) {
      x = min(d / i, n - c);
      x = min(x, 1 << i);
      a[i] = x;
      c += x;
      d -= (x * i);
    }
    if (c < n) {
      cout << "NO" << endl;
      continue;
    }
    flag = 0;
    while (d--) {
      f = 0;
      for (int i = 2; i <= 100; ++i) {
        if (a[i - 1] > 1 && 2 * a[i - 1] >= a[i] + 3) {
          a[i - 1]--;
          a[i]++;
          f = 1;
          break;
        }
      }
      if (!f) {
        flag = 1;
        break;
      }
    }
    if (flag) {
      cout << "NO" << endl;
      continue;
    }
    cout << "YES" << endl;
    vector<int> v;
    v.push_back(1);
    c = 1;
    for (int i = 1; i <= 100; ++i) {
      vector<int> g;
      for (int j = 1; j <= a[i]; ++j) {
        ++c;
        g.push_back(c);
        cout << v[(j - 1) / 2] << " ";
      }
      v = g;
    }
    cout << endl;
  }
}