#include <bits/stdc++.h>
using namespace std;
template <typename T>
istream& operator>>(istream& is, vector<T>& a) {
  for (T& ai : a) is >> ai;
  return is;
}
template <typename T>
ostream& operator<<(ostream& os, vector<T> const& a) {
  os << "[ ";
  for (const T& ai : a) os << ai << " ";
  return os << "]";
}
template <typename T1, typename T2>
istream& operator>>(istream& is, pair<T1, T2>& a) {
  return is >> a.first >> a.second;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& os, pair<T1, T2> const& a) {
  return os << "(" << a.first << ", " << a.second << ")";
}
void f() {
  int n, d;
  cin >> n >> d;
  if (d > n * (n - 1) / 2) {
    cout << "NO\n";
    return;
  }
  vector<int> c;
  int x = n;
  for (int i = 0; x; ++i) {
    c.push_back(min(x, 1 << i));
    d -= i * c.back();
    x -= c.back();
  }
  if (d < 0) {
    cout << "NO\n";
    return;
  }
  int t = 0;
  while (d) {
    if (!t) {
      t = c.size() - 1;
      if (c.back())
        c.push_back(0);
      else
        --t;
      continue;
    }
    int k = min((2 * c[t] - c[t + 1]) / 3, d);
    d -= k;
    c[t + 1] += k;
    c[t--] -= k;
  }
  cout << "YES\n";
  int r = 1;
  for (int i = 1, h = c.size(); i < h; ++i) {
    for (int j = 0; j < c[i]; ++j) {
      cout << r + (j >> 1) << " ";
    }
    r += c[i - 1];
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) f();
  return 0;
}

### Prompt

Generate 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;
template <typename T>
istream& operator>>(istream& is, vector<T>& a) {
  for (T& ai : a) is >> ai;
  return is;
}
template <typename T>
ostream& operator<<(ostream& os, vector<T> const& a) {
  os << "[ ";
  for (const T& ai : a) os << ai << " ";
  return os << "]";
}
template <typename T1, typename T2>
istream& operator>>(istream& is, pair<T1, T2>& a) {
  return is >> a.first >> a.second;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& os, pair<T1, T2> const& a) {
  return os << "(" << a.first << ", " << a.second << ")";
}
void f() {
  int n, d;
  cin >> n >> d;
  if (d > n * (n - 1) / 2) {
    cout << "NO\n";
    return;
  }
  vector<int> c;
  int x = n;
  for (int i = 0; x; ++i) {
    c.push_back(min(x, 1 << i));
    d -= i * c.back();
    x -= c.back();
  }
  if (d < 0) {
    cout << "NO\n";
    return;
  }
  int t = 0;
  while (d) {
    if (!t) {
      t = c.size() - 1;
      if (c.back())
        c.push_back(0);
      else
        --t;
      continue;
    }
    int k = min((2 * c[t] - c[t + 1]) / 3, d);
    d -= k;
    c[t + 1] += k;
    c[t--] -= k;
  }
  cout << "YES\n";
  int r = 1;
  for (int i = 1, h = c.size(); i < h; ++i) {
    for (int j = 0; j < c[i]; ++j) {
      cout << r + (j >> 1) << " ";
    }
    r += c[i - 1];
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) f();
  return 0;
}

```