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:

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#include <bits/stdc++.h>
using namespace std;
int mi(int n) {
  int r = 0;
  for (int i = 0; n; i++) {
    int take = min(1 << i, n);
    n -= take;
    r += i * take;
  }
  return r;
}
int ma(int n) { return n * (n - 1) / 2; }
int id;
void build(int n, int d, int par) {
  assert(d >= mi(n) && d <= ma(n));
  if (n == 0) {
    assert(d == 0);
    return;
  }
  d -= n - 1;
  if (par != -1) cout << par << ' ';
  int cur = id++;
  for (int i = 0; i <= n; i++) {
    if (mi(i) + mi(n - 1 - i) <= d && ma(i) + ma(n - 1 - i) >= d) {
      int a = min(d - mi(n - 1 - i), ma(i));
      build(i, a, cur);
      build(n - 1 - i, d - a, cur);
      return;
    }
  }
  assert(0);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n, d;
    cin >> n >> d;
    if (mi(n) > d || ma(n) < d) {
      cout << "NO" << endl;
      continue;
    }
    id = 1;
    cout << "YES" << endl;
    build(n, d, -1);
    cout << endl;
  }
}
