#include <bits/stdc++.h>
using namespace std;
int t, n, d;
int par[5002];
int main() {
  cin >> t;
  for (int cas = 1; cas <= t; ++cas) {
    cin >> n >> d;
    long long maxv = (long long)n * (long long)(n - 1) / 2;
    long long minv = 0LL;
    int lev = 0;
    int tot = 1;
    int cur = 1;
    while (true) {
      cur *= 2;
      lev++;
      if (tot + cur <= n) {
        tot += cur;
        minv += cur * lev;
      } else {
        break;
      }
    }
    minv += (n - tot) * lev;
    if (d < minv || d > maxv) {
      cout << "NO" << endl;
    } else {
      cout << "YES" << endl;
      unordered_map<int, vector<int>> pts;
      pts.clear();
      for (int i = 1; i <= n; ++i) {
        pts[i - 1].push_back(i);
        if (i > 1) par[i] = i - 1;
      }
      int tot = maxv;
      int cl = n - 1;
      int canl = 0;
      int canp = 0;
      while (tot > d) {
        int x = pts[cl][0];
        int det = tot - d;
        if (det >= cl - canl - 1) {
          tot -= cl - canl - 1;
          pts[cl].pop_back();
          pts[canl + 1].push_back(x);
          par[x] = pts[canl][canp];
          if (pts[canl + 1].size() == (1 << (canl + 1))) {
            canl++;
            canp = 0;
          } else {
            canp = (pts[canl + 1].size()) / 2;
          }
        } else {
          pts[cl].pop_back();
          pts[cl - det].push_back(x);
          par[x] = pts[cl - det - 1][0];
          break;
        }
        cl--;
      }
      for (int i = 2; i <= n; ++i) {
        cout << par[i];
        if (i < n)
          cout << " ";
        else
          cout << endl;
      }
    }
  }
  return 0;
}

### Prompt

Please formulate 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 t, n, d;
int par[5002];
int main() {
  cin >> t;
  for (int cas = 1; cas <= t; ++cas) {
    cin >> n >> d;
    long long maxv = (long long)n * (long long)(n - 1) / 2;
    long long minv = 0LL;
    int lev = 0;
    int tot = 1;
    int cur = 1;
    while (true) {
      cur *= 2;
      lev++;
      if (tot + cur <= n) {
        tot += cur;
        minv += cur * lev;
      } else {
        break;
      }
    }
    minv += (n - tot) * lev;
    if (d < minv || d > maxv) {
      cout << "NO" << endl;
    } else {
      cout << "YES" << endl;
      unordered_map<int, vector<int>> pts;
      pts.clear();
      for (int i = 1; i <= n; ++i) {
        pts[i - 1].push_back(i);
        if (i > 1) par[i] = i - 1;
      }
      int tot = maxv;
      int cl = n - 1;
      int canl = 0;
      int canp = 0;
      while (tot > d) {
        int x = pts[cl][0];
        int det = tot - d;
        if (det >= cl - canl - 1) {
          tot -= cl - canl - 1;
          pts[cl].pop_back();
          pts[canl + 1].push_back(x);
          par[x] = pts[canl][canp];
          if (pts[canl + 1].size() == (1 << (canl + 1))) {
            canl++;
            canp = 0;
          } else {
            canp = (pts[canl + 1].size()) / 2;
          }
        } else {
          pts[cl].pop_back();
          pts[cl - det].push_back(x);
          par[x] = pts[cl - det - 1][0];
          break;
        }
        cl--;
      }
      for (int i = 2; i <= n; ++i) {
        cout << par[i];
        if (i < n)
          cout << " ";
        else
          cout << endl;
      }
    }
  }
  return 0;
}

```