#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cout.tie(nullptr);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    int n, d, cur;
    cin >> n >> d;
    cur = n * (n - 1) / 2;
    if (d > cur) {
      cout << "NO\n";
      continue;
    }
    vector<int> p(n), dep(n), child(n, 1);
    iota(p.begin() + 1, p.end(), 0);
    iota(dep.begin(), dep.end(), 0);
    child[n - 1] = 0;
    for (int i = n - 1; i > 0; --i) {
      int par = -1;
      for (int j = n - 1; j >= 0; --j) {
        if (i == j || child[j] == 2 || (cur - dep[i] + dep[j] + 1 < d) ||
            dep[i] <= dep[j] + 1)
          continue;
        if (par == -1 || dep[j] < dep[par]) par = j;
      }
      if (par != -1) {
        p[i] = par;
        child[par]++;
        cur -= dep[i] - dep[par] - 1;
        dep[i] = dep[par] + 1;
        child[i] = 0;
      }
    }
    if (cur != d)
      cout << "NO\n";
    else {
      cout << "YES\n";
      for (int i = 1; i < n; ++i) {
        cout << p[i] + 1 << " ";
      }
      cout << '\n';
    }
  }
}

### Prompt

Generate a CPP solution to the following 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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cout.tie(nullptr);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    int n, d, cur;
    cin >> n >> d;
    cur = n * (n - 1) / 2;
    if (d > cur) {
      cout << "NO\n";
      continue;
    }
    vector<int> p(n), dep(n), child(n, 1);
    iota(p.begin() + 1, p.end(), 0);
    iota(dep.begin(), dep.end(), 0);
    child[n - 1] = 0;
    for (int i = n - 1; i > 0; --i) {
      int par = -1;
      for (int j = n - 1; j >= 0; --j) {
        if (i == j || child[j] == 2 || (cur - dep[i] + dep[j] + 1 < d) ||
            dep[i] <= dep[j] + 1)
          continue;
        if (par == -1 || dep[j] < dep[par]) par = j;
      }
      if (par != -1) {
        p[i] = par;
        child[par]++;
        cur -= dep[i] - dep[par] - 1;
        dep[i] = dep[par] + 1;
        child[i] = 0;
      }
    }
    if (cur != d)
      cout << "NO\n";
    else {
      cout << "YES\n";
      for (int i = 1; i < n; ++i) {
        cout << p[i] + 1 << " ";
      }
      cout << '\n';
    }
  }
}

```