#include <bits/stdc++.h>
using namespace std;
const int N = 5e3 + 10;
int p[N], ln[N], pre[N], c[N], d[N];
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  for (int i = 2; i < N; ++i) {
    ln[i] = ln[i >> 1] + 1;
    pre[i] = pre[i - 1] + ln[i];
  }
  while (t--) {
    int n, z;
    cin >> n >> z;
    if (((n * (n - 1)) >> 1) < z) {
      cout << "NO\n";
      continue;
    }
    int sum = pre[n];
    if (sum > z) {
      cout << "NO\n";
      continue;
    }
    set<int> leaves;
    set<pair<int, int>> s;
    for (int i = 1; i <= n; ++i) {
      c[i] = 0;
      p[i] = i >> 1;
      d[i] = ln[i];
      if ((i << 1 | 1) > n) {
        s.insert({d[i], i});
        if ((i << 1) > n)
          leaves.insert(i);
        else
          c[i] = 1;
      } else
        c[i] = 2;
    }
    while (sum < z) {
      for (auto &i : leaves) {
        auto it = s.lower_bound({d[i], -1});
        if (it != s.end()) {
          if (it->second == i) ++it;
          if (it != s.end()) {
            if (it->first != d[i]) continue;
            ++sum;
            ++c[it->second];
            --c[p[i]];
            auto x = s.lower_bound({d[i], i});
            if (x != s.end()) s.erase(x);
            ++d[i];
            s.insert({d[i], i});
            if (!c[p[i]])
              leaves.insert(p[i]);
            else
              s.insert({d[p[i]], p[i]});
            p[i] = it->second;
            if (c[it->second] == 2)
              s.erase(it);
            else
              leaves.erase(leaves.find(it->second));
            break;
          }
        }
      }
    }
    if (sum < z) {
      cout << "NO";
      continue;
    } else {
      cout << "YES\n";
      for (int i = 2; i <= n; ++i) cout << p[i] << ' ';
    }
    cout << '\n';
  }
  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;
const int N = 5e3 + 10;
int p[N], ln[N], pre[N], c[N], d[N];
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  for (int i = 2; i < N; ++i) {
    ln[i] = ln[i >> 1] + 1;
    pre[i] = pre[i - 1] + ln[i];
  }
  while (t--) {
    int n, z;
    cin >> n >> z;
    if (((n * (n - 1)) >> 1) < z) {
      cout << "NO\n";
      continue;
    }
    int sum = pre[n];
    if (sum > z) {
      cout << "NO\n";
      continue;
    }
    set<int> leaves;
    set<pair<int, int>> s;
    for (int i = 1; i <= n; ++i) {
      c[i] = 0;
      p[i] = i >> 1;
      d[i] = ln[i];
      if ((i << 1 | 1) > n) {
        s.insert({d[i], i});
        if ((i << 1) > n)
          leaves.insert(i);
        else
          c[i] = 1;
      } else
        c[i] = 2;
    }
    while (sum < z) {
      for (auto &i : leaves) {
        auto it = s.lower_bound({d[i], -1});
        if (it != s.end()) {
          if (it->second == i) ++it;
          if (it != s.end()) {
            if (it->first != d[i]) continue;
            ++sum;
            ++c[it->second];
            --c[p[i]];
            auto x = s.lower_bound({d[i], i});
            if (x != s.end()) s.erase(x);
            ++d[i];
            s.insert({d[i], i});
            if (!c[p[i]])
              leaves.insert(p[i]);
            else
              s.insert({d[p[i]], p[i]});
            p[i] = it->second;
            if (c[it->second] == 2)
              s.erase(it);
            else
              leaves.erase(leaves.find(it->second));
            break;
          }
        }
      }
    }
    if (sum < z) {
      cout << "NO";
      continue;
    } else {
      cout << "YES\n";
      for (int i = 2; i <= n; ++i) cout << p[i] << ' ';
    }
    cout << '\n';
  }
  return 0;
}

```