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 n, d;
const int N = 5e3 + 47, P = 31;
int cnt[N], ans[N], P2[P + 1];
void gen() {
  P2[0] = 1;
  for (int i = 1; i <= P; i++) P2[i] = P2[i - 1] * 2;
}
void setup() {
  for (int i = 1; i <= n; i++) cnt[i] = 1;
}
int gen_ans(int id, int parent, int row) {
  ans[id] = parent;
  if (cnt[row + 1] == 0) return id + 1;
  if (cnt[row + 1] == 1) {
    cnt[row + 1]--;
    return gen_ans(id + 1, id, row + 1);
  }
  if (cnt[row + 1] >= 2) {
    cnt[row + 1] -= 2;
    int old_id = id;
    id = gen_ans(id + 1, old_id, row + 1);
    return gen_ans(id, old_id, row + 1);
  }
}
bool solvable() {
  int sum = (n * (n - 1)) / 2;
  if (sum < d) return false;
  int upper_row = 2, lower_row = n;
  while (sum != d) {
    if (lower_row <= upper_row) break;
    if ((lower_row - upper_row) <= (sum - d)) {
      cnt[upper_row]++;
      cnt[lower_row]--;
      sum -= (lower_row - upper_row);
      if (cnt[upper_row] == P2[upper_row - 1]) upper_row++;
      if (cnt[lower_row] == 0) lower_row--;
    } else {
      cnt[lower_row - sum + d]++;
      cnt[lower_row]--;
      sum -= (sum - d);
    }
  }
  if (sum == d) return true;
  return false;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  gen();
  int T;
  cin >> T;
  while (T--) {
    cin >> n >> d;
    setup();
    if (solvable()) {
      cout << "YES\n";
      gen_ans(1, 1, 1);
      for (int i = 2; i <= n; i++) {
        cout << ans[i] << " ";
      }
      cout << "\n";
    } else
      cout << "NO\n";
  }
}