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;
int max2pow(int n) {
  int result = 2, i;
  for (i = 0; result <= n; ++i, result <<= 1)
    ;
  return i;
}
void solve() {
  int n, d;
  cin >> n >> d;
  if (d > n * (n - 1) / 2 ||
      d < (1 << max2pow(n + 1)) * (max2pow(n + 1) - 2) + 2 +
              (max2pow(n + 1) * (n - (1 << max2pow(n + 1)) + 1))) {
    cout << "NO\n";
    return;
  }
  cout << "YES\n";
  vector<int> level(n, 1);
  int lastlevel = n - 2, currentlevel = n - 1, fillinglevel = 1;
  for (int sum = n * (n - 1) / 2; sum > d; --sum) {
    if (currentlevel - 1 == fillinglevel &&
        level[fillinglevel] == (1 << fillinglevel)) {
      ++fillinglevel;
      currentlevel = lastlevel;
      --lastlevel;
    }
    --level[currentlevel];
    --currentlevel;
    ++level[currentlevel];
    if (currentlevel == fillinglevel) {
      currentlevel = lastlevel;
      --lastlevel;
    }
  }
  int current_level = 1;
  int element = 2;
  vector<int> v1;
  vector<int> v2;
  vector<int> *prev_level = &v1;
  vector<int> *next_level = &v2;
  prev_level->push_back(1);
  while (current_level < n && level[current_level] > 0) {
    for (int i = 0; i < level[current_level]; ++i) {
      cout << prev_level->at(i / 2) << ' ';
      next_level->push_back(element);
      ++element;
    }
    ++current_level;
    prev_level->clear();
    swap(prev_level, next_level);
  }
  cout << '\n';
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
