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 is_power_of_2(int n) { return ((n & (n - 1)) == 0); }
void solve() {
  int no_of_vertices, total_depth_sum;
  cin >> no_of_vertices >> total_depth_sum;
  long long maximum_sum = (no_of_vertices * (no_of_vertices - 1)) / 2;
  long long minimum_sum = 0;
  for (int i = 1, level = 0; i <= no_of_vertices; i++) {
    if (is_power_of_2(i)) {
      level++;
    }
    minimum_sum += (level - 1);
  }
  if (total_depth_sum < minimum_sum || maximum_sum < total_depth_sum) {
    cout << "NO\n";
    return;
  }
  vector<int> depth(no_of_vertices + 1, 0);
  vector<int> parent(no_of_vertices + 1, 0);
  vector<int> no_of_children(no_of_vertices + 1, 1);
  for (int i = 1; i <= no_of_vertices; i++) {
    parent[i] = i - 1;
    depth[i] = i - 1;
  }
  no_of_children[no_of_vertices] = 0;
  vector<int> immobile(no_of_vertices + 1, false);
  for (long long sum = maximum_sum; sum > total_depth_sum;) {
    int first_v = 0;
    for (int i = 1; i <= no_of_vertices; i++) {
      if (!immobile[i] && no_of_children[i] == 0 &&
          (first_v == 0 || depth[i] < depth[first_v])) {
        first_v = i;
      }
    }
    int best_parent = 0;
    for (int i = 1; i <= no_of_vertices; i++) {
      if (depth[i] == depth[first_v] - 2 && no_of_children[i] < 2) {
        best_parent = i;
      }
    }
    if (best_parent == 0) {
      immobile[first_v] = true;
      continue;
    }
    sum--;
    no_of_children[parent[first_v]]--;
    depth[first_v]--;
    parent[first_v] = best_parent;
    no_of_children[best_parent]++;
  }
  cout << "YES\n";
  for (int i = 2; i <= no_of_vertices; i++) {
    cout << parent[i] << " ";
  }
  cout << "\n";
}
int main() {
  int no_of_test_cases;
  cin >> no_of_test_cases;
  while (no_of_test_cases--) solve();
  return 0;
}