#include <bits/stdc++.h>
using namespace std;
void solve(const int& n, int d, ostream& cout) {
  vector<vector<int>> nodes_from(n, vector<int>());
  vector<int> p(n + 1);
  int min_sum = 0;
  int max_sum = ((n - 1) * n) / 2;
  int level = 1;
  int level_max_node = 3;
  for (int i = 2; i <= n; ++i) {
    if (level_max_node < i) {
      level_max_node = 2 * level_max_node + 1;
      level += 1;
    }
    p[i] = i / 2;
    min_sum += level;
    nodes_from[level].push_back(i);
  }
  if (min_sum <= d && d <= max_sum) {
    int level = 1;
    while (level + 1 < n && nodes_from[level + 1].size() > 0) {
      ++level;
    }
    int sum = min_sum;
    while (sum < d) {
      while (nodes_from[level].size() <= 1) {
        --level;
      }
      int node = nodes_from[level].back();
      int new_parent =
          nodes_from[level].front() + nodes_from[level + 1].size() / 2;
      p[node] = new_parent;
      nodes_from[level + 1].push_back(node);
      nodes_from[level].pop_back();
      ++level;
      ++sum;
    }
    cout << "YES\n";
    for (int i = 2; i <= n; ++i) {
      cout << p[i] << " ";
    }
    cout << "\n";
  } else {
    cout << "NO\n";
  }
}
int main(int argc, char** argv) {
  int t;
  cin >> t;
  for (; t > 0; --t) {
    int n, d;
    cin >> n >> d;
    solve(n, d, cout);
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
void solve(const int& n, int d, ostream& cout) {
  vector<vector<int>> nodes_from(n, vector<int>());
  vector<int> p(n + 1);
  int min_sum = 0;
  int max_sum = ((n - 1) * n) / 2;
  int level = 1;
  int level_max_node = 3;
  for (int i = 2; i <= n; ++i) {
    if (level_max_node < i) {
      level_max_node = 2 * level_max_node + 1;
      level += 1;
    }
    p[i] = i / 2;
    min_sum += level;
    nodes_from[level].push_back(i);
  }
  if (min_sum <= d && d <= max_sum) {
    int level = 1;
    while (level + 1 < n && nodes_from[level + 1].size() > 0) {
      ++level;
    }
    int sum = min_sum;
    while (sum < d) {
      while (nodes_from[level].size() <= 1) {
        --level;
      }
      int node = nodes_from[level].back();
      int new_parent =
          nodes_from[level].front() + nodes_from[level + 1].size() / 2;
      p[node] = new_parent;
      nodes_from[level + 1].push_back(node);
      nodes_from[level].pop_back();
      ++level;
      ++sum;
    }
    cout << "YES\n";
    for (int i = 2; i <= n; ++i) {
      cout << p[i] << " ";
    }
    cout << "\n";
  } else {
    cout << "NO\n";
  }
}
int main(int argc, char** argv) {
  int t;
  cin >> t;
  for (; t > 0; --t) {
    int n, d;
    cin >> n >> d;
    solve(n, d, cout);
  }
  return 0;
}

```