#include <bits/stdc++.h>
using namespace std;
int n, d;
void solve() {
  cin >> n >> d;
  if (n * (n - 1) / 2 < d) {
    cout << "NO\n";
    return;
  }
  vector<int> lvl(n + 1, 0), pr(n + 1, 0);
  vector<vector<int>> last(n + 1);
  ;
  lvl[0] = 1;
  lvl[1] = 1;
  last[0].push_back(1);
  for (int i = 2; i <= n; i++) {
    lvl[i] = 1;
    pr[i] = i - 1;
    last[i - 1].push_back(i);
  }
  int cur = n * (n - 1) / 2;
  int le = n - 1;
  int l = le;
  while (cur > d) {
    int node = last[le][0];
    l = le;
    bool f = 0;
    l--;
    while (cur > d && lvl[l] < pow(2, l)) {
      f = 1;
      lvl[l]++;
      lvl[l + 1]--;
      last[l].push_back(node);
      pr[node] = last[l - 1][((lvl[l] + 1) / 2) - 1];
      last[l + 1].pop_back();
      l--;
      cur--;
    }
    if (!f) break;
    le--;
  }
  if (cur > d)
    cout << "NO\n";
  else {
    cout << "YES\n";
    for (int i = 2; i <= n; i++) cout << pr[i] << " \n"[i == n];
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  ;
  int T;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int n, d;
void solve() {
  cin >> n >> d;
  if (n * (n - 1) / 2 < d) {
    cout << "NO\n";
    return;
  }
  vector<int> lvl(n + 1, 0), pr(n + 1, 0);
  vector<vector<int>> last(n + 1);
  ;
  lvl[0] = 1;
  lvl[1] = 1;
  last[0].push_back(1);
  for (int i = 2; i <= n; i++) {
    lvl[i] = 1;
    pr[i] = i - 1;
    last[i - 1].push_back(i);
  }
  int cur = n * (n - 1) / 2;
  int le = n - 1;
  int l = le;
  while (cur > d) {
    int node = last[le][0];
    l = le;
    bool f = 0;
    l--;
    while (cur > d && lvl[l] < pow(2, l)) {
      f = 1;
      lvl[l]++;
      lvl[l + 1]--;
      last[l].push_back(node);
      pr[node] = last[l - 1][((lvl[l] + 1) / 2) - 1];
      last[l + 1].pop_back();
      l--;
      cur--;
    }
    if (!f) break;
    le--;
  }
  if (cur > d)
    cout << "NO\n";
  else {
    cout << "YES\n";
    for (int i = 2; i <= n; i++) cout << pr[i] << " \n"[i == n];
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  ;
  int T;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}

```