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 N = 5005;
int cnt[N];
inline int getMin(int n, int r) {
  int ret = 0, cur = 0;
  while (n) {
    int tmp = min(n, r);
    ret += tmp * cur;
    n -= tmp;
    r <<= 1;
    cur++;
  }
  return ret;
}
inline int getMax(int n, int r) {
  n -= r;
  return n * (n + 1) / 2;
}
void process(int ind, int n, int d) {
  if (d == 0) {
    cnt[ind] = n;
    return;
  }
  int r = cnt[ind - 1];
  for (int i = 1; i <= 2 * r && i <= n; i++) {
    if (getMin(n, i) <= d && d <= getMax(n, i)) {
      cnt[ind] = i;
      process(ind + 1, n - i, d - n + i);
      return;
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << setprecision(32);
  int t;
  cin >> t;
  while (t--) {
    int n, d;
    cin >> n >> d;
    if (d < getMin(n, 1) || d > getMax(n, 1)) {
      cout << "NO" << '\n';
      continue;
    }
    cout << "YES" << '\n';
    memset(cnt, 0, sizeof(cnt));
    cnt[0] = 1;
    process(1, n - 1, d - n + 1);
    vector<int> vec({1});
    int ind = 1;
    for (int i = 1; ind < n; i++) {
      vector<int> nvec;
      for (int j = 0; j < cnt[i]; j++) {
        nvec.push_back(++ind);
        cout << vec[j / 2] << " ";
      }
      vec = nvec;
    }
    cout << '\n';
  }
  return 0;
}