#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    int n, d;
    cin >> n >> d;
    int D = (int)(ceil(log2(n + 1))) - 1;
    int minsd = (n + 1) * D - (1 << (D + 1)) + 2;
    if (d < minsd || d > (n * (n - 1) / 2)) {
      cout << "NO\n";
      continue;
    }
    cout << "YES\n";
    int *dep = new int[n + 1], p[n + 1], c[n + 1];
    vector<int> v;
    for (int i = 0; i < n + 1; i++) {
      v.push_back(i);
      c[i] = 0;
    }
    for (int i = 0; i < n + 1; i++) {
      p[i] = (i >> 1);
      if (2 * i <= n) c[i]++;
      if (2 * i + 1 <= n) c[i]++;
      dep[i] = (int)ceil(log2(i + 1)) - 1;
    }
    auto fun = [dep](int a, int b) { return dep[a] < dep[b]; };
    for (int sd = minsd + 1; sd <= d && sd <= n * (n - 1) / 2; sd++) {
      sort(v.begin(), v.end(), fun);
      int a = -1, b = -1, i = 1;
      while (i <= n) {
        int a1 = -1, b1 = -1;
        int de = dep[v[i]];
        while (i <= n && dep[v[i]] == de) {
          if (a1 == -1 && c[v[i]] == 0)
            a1 = v[i];
          else if (a1 != -1 && c[v[i]] < 2)
            b1 = v[i];
          if (b1 == -1 && c[v[i]] == 1) b1 = v[i];
          if (a1 != -1 && b1 != -1) break;
          i++;
        }
        if (a1 != -1 && b1 != -1) {
          a = a1;
          b = b1;
          break;
        }
      }
      c[p[a]]--;
      p[a] = b;
      c[b]++;
      dep[a]++;
    }
    for (int i = 2; i <= n; i++) cout << p[i] << " ";
    cout << endl;
  }
}

### 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 main() {
  ios_base::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    int n, d;
    cin >> n >> d;
    int D = (int)(ceil(log2(n + 1))) - 1;
    int minsd = (n + 1) * D - (1 << (D + 1)) + 2;
    if (d < minsd || d > (n * (n - 1) / 2)) {
      cout << "NO\n";
      continue;
    }
    cout << "YES\n";
    int *dep = new int[n + 1], p[n + 1], c[n + 1];
    vector<int> v;
    for (int i = 0; i < n + 1; i++) {
      v.push_back(i);
      c[i] = 0;
    }
    for (int i = 0; i < n + 1; i++) {
      p[i] = (i >> 1);
      if (2 * i <= n) c[i]++;
      if (2 * i + 1 <= n) c[i]++;
      dep[i] = (int)ceil(log2(i + 1)) - 1;
    }
    auto fun = [dep](int a, int b) { return dep[a] < dep[b]; };
    for (int sd = minsd + 1; sd <= d && sd <= n * (n - 1) / 2; sd++) {
      sort(v.begin(), v.end(), fun);
      int a = -1, b = -1, i = 1;
      while (i <= n) {
        int a1 = -1, b1 = -1;
        int de = dep[v[i]];
        while (i <= n && dep[v[i]] == de) {
          if (a1 == -1 && c[v[i]] == 0)
            a1 = v[i];
          else if (a1 != -1 && c[v[i]] < 2)
            b1 = v[i];
          if (b1 == -1 && c[v[i]] == 1) b1 = v[i];
          if (a1 != -1 && b1 != -1) break;
          i++;
        }
        if (a1 != -1 && b1 != -1) {
          a = a1;
          b = b1;
          break;
        }
      }
      c[p[a]]--;
      p[a] = b;
      c[b]++;
      dep[a]++;
    }
    for (int i = 2; i <= n; i++) cout << p[i] << " ";
    cout << endl;
  }
}

```