#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, d;
    cin >> n >> d;
    vector<int> v(0);
    int mi = 0, ma = (n * (n - 1)) / 2;
    int i = 0;
    while (n > 0) {
      mi += min(n, (int)pow(2, i)) * i;
      v.push_back(min(n, (int)pow(2, i)));
      n -= min(n, (int)pow(2, i));
      i++;
    }
    if (d < mi || d > ma)
      cout << "NO" << endl;
    else {
      int soma = mi, tam = v.size();
      for (int i = tam - 1; soma != d && i > 0; i--) {
        while (v[i] > 1 && soma != d) {
          if (d - soma >= v.size() - i) {
            soma += v.size() - i;
            v.push_back(1);
          } else {
            v[i + d - soma]++;
            soma = d;
          }
          v[i]--;
        }
      }
      cout << "YES" << endl;
      vector<vector<int> > m(v.size(), vector<int>(0));
      int nod = 1;
      for (int i = 0; i < v.size(); i++) {
        for (int j = nod; j < nod + v[i]; j++) {
          m[i].push_back(j);
        }
        nod += v[i];
      }
      for (int i = 1; i < v.size(); i++) {
        for (int j = 0; j < m[i].size(); j++) {
          cout << m[i - 1][j / 2] << " ";
        }
      }
      cout << endl;
    }
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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() {
  int t;
  cin >> t;
  while (t--) {
    int n, d;
    cin >> n >> d;
    vector<int> v(0);
    int mi = 0, ma = (n * (n - 1)) / 2;
    int i = 0;
    while (n > 0) {
      mi += min(n, (int)pow(2, i)) * i;
      v.push_back(min(n, (int)pow(2, i)));
      n -= min(n, (int)pow(2, i));
      i++;
    }
    if (d < mi || d > ma)
      cout << "NO" << endl;
    else {
      int soma = mi, tam = v.size();
      for (int i = tam - 1; soma != d && i > 0; i--) {
        while (v[i] > 1 && soma != d) {
          if (d - soma >= v.size() - i) {
            soma += v.size() - i;
            v.push_back(1);
          } else {
            v[i + d - soma]++;
            soma = d;
          }
          v[i]--;
        }
      }
      cout << "YES" << endl;
      vector<vector<int> > m(v.size(), vector<int>(0));
      int nod = 1;
      for (int i = 0; i < v.size(); i++) {
        for (int j = nod; j < nod + v[i]; j++) {
          m[i].push_back(j);
        }
        nod += v[i];
      }
      for (int i = 1; i < v.size(); i++) {
        for (int j = 0; j < m[i].size(); j++) {
          cout << m[i - 1][j / 2] << " ";
        }
      }
      cout << endl;
    }
  }
  return 0;
}

```