#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n, d;
    cin >> n >> d;
    long long x = n * (n - 1) / 2;
    if (d > x) {
      cout << "NO\n";
      continue;
    }
    x = 1;
    long long s = 0;
    vector<long long> a;
    for (long long i = 0; i < n; i++) {
      for (long long j = 0; j < x; j++) {
        if (s + i > d or a.size() == n) break;
        s += i;
        a.push_back(i);
      }
      x *= 2;
      if (s + i > d or a.size() == n) break;
    }
    if (a.size() < n) {
      cout << "NO\n";
      continue;
    }
    cout << "YES\n";
    d -= s;
    long long p = a.size();
    for (long long i = 0; i < d; i++) {
      for (long long j = p - 1; j > 0; j--) {
        if (a[j] == a[j - 1]) {
          a[j] += 1;
          break;
        }
      }
    }
    long long j = 0, pp = 0;
    vector<long long> b;
    for (long long i = 1; i < p; i++) {
      while (a[j] != a[i] - 1) {
        j++;
        pp = 0;
        continue;
      }
      if (pp == 0) {
        b.push_back(j + 1);
        pp = 1;
      } else {
        pp = 0;
        b.push_back(j + 1);
        j += 1;
      }
    }
    for (auto i : b) {
      cout << i << " ";
    }
    cout << "\n";
  }
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n, d;
    cin >> n >> d;
    long long x = n * (n - 1) / 2;
    if (d > x) {
      cout << "NO\n";
      continue;
    }
    x = 1;
    long long s = 0;
    vector<long long> a;
    for (long long i = 0; i < n; i++) {
      for (long long j = 0; j < x; j++) {
        if (s + i > d or a.size() == n) break;
        s += i;
        a.push_back(i);
      }
      x *= 2;
      if (s + i > d or a.size() == n) break;
    }
    if (a.size() < n) {
      cout << "NO\n";
      continue;
    }
    cout << "YES\n";
    d -= s;
    long long p = a.size();
    for (long long i = 0; i < d; i++) {
      for (long long j = p - 1; j > 0; j--) {
        if (a[j] == a[j - 1]) {
          a[j] += 1;
          break;
        }
      }
    }
    long long j = 0, pp = 0;
    vector<long long> b;
    for (long long i = 1; i < p; i++) {
      while (a[j] != a[i] - 1) {
        j++;
        pp = 0;
        continue;
      }
      if (pp == 0) {
        b.push_back(j + 1);
        pp = 1;
      } else {
        pp = 0;
        b.push_back(j + 1);
        j += 1;
      }
    }
    for (auto i : b) {
      cout << i << " ";
    }
    cout << "\n";
  }
}

```