#include <bits/stdc++.h>
using lint = long long;
using ldouble = long double;
void solve() {
  int n, d;
  std::cin >> n >> d;
  std::vector<int> vs(n, 1);
  int diff = n * (n - 1) / 2 - d;
  if (diff < 0) {
    std::cout << "NO" << std::endl;
    return;
  }
  int head = 1;
  for (int v = n - 1; v > head; --v) {
    int to = std::max(v - diff, head);
    --vs[v];
    ++vs[to];
    diff -= v - to;
    if (diff == 0) break;
    while (vs[head] == (1 << head)) ++head;
  }
  if (diff > 0) {
    std::cout << "NO" << std::endl;
    return;
  }
  std::cout << "YES" << std::endl;
  int p = 1;
  for (int i = 1; i < n; ++i) {
    int v = p, ch = 0;
    for (int j = 0; j < vs[i]; ++j) {
      std::cout << v << " ";
      if (++ch == 2) {
        ++v;
        ch = 0;
      }
    }
    p += vs[i - 1];
  }
  std::cout << std::endl;
}
int main() {
  std::cin.tie(nullptr);
  std::cout.tie(nullptr);
  std::ios::sync_with_stdio(false);
  int q;
  std::cin >> q;
  while (q--) solve();
  return 0;
}

### Prompt

Your challenge is to write 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 lint = long long;
using ldouble = long double;
void solve() {
  int n, d;
  std::cin >> n >> d;
  std::vector<int> vs(n, 1);
  int diff = n * (n - 1) / 2 - d;
  if (diff < 0) {
    std::cout << "NO" << std::endl;
    return;
  }
  int head = 1;
  for (int v = n - 1; v > head; --v) {
    int to = std::max(v - diff, head);
    --vs[v];
    ++vs[to];
    diff -= v - to;
    if (diff == 0) break;
    while (vs[head] == (1 << head)) ++head;
  }
  if (diff > 0) {
    std::cout << "NO" << std::endl;
    return;
  }
  std::cout << "YES" << std::endl;
  int p = 1;
  for (int i = 1; i < n; ++i) {
    int v = p, ch = 0;
    for (int j = 0; j < vs[i]; ++j) {
      std::cout << v << " ";
      if (++ch == 2) {
        ++v;
        ch = 0;
      }
    }
    p += vs[i - 1];
  }
  std::cout << std::endl;
}
int main() {
  std::cin.tie(nullptr);
  std::cout.tie(nullptr);
  std::ios::sync_with_stdio(false);
  int q;
  std::cin >> q;
  while (q--) solve();
  return 0;
}

```