#include <bits/stdc++.h>
using namespace std;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  for (; t--;) {
    int n, d;
    cin >> n >> d;
    int Min = 0, Max = n * (n - 1) / 2, now = 0;
    for (int i = 2; i <= n; ++i) {
      if ((1 << (now + 1)) == i) ++now;
      Min += now;
    }
    if (d < Min || Max < d) {
      cout << "NO" << '\n';
      continue;
    }
    cout << "YES" << '\n';
    d = Max - d;
    vector<int> par(n), dep(n), deg(n, 1);
    iota(par.begin(), par.end(), -1);
    iota(dep.begin(), dep.end(), 0);
    deg[n - 1] = 0;
    for (int i = n - 1; i > 0; --i) {
      if (!d) break;
      int m = n, node;
      for (int j = 0; j < n; ++j)
        if (j != i && deg[j] != 2) {
          if (dep[i] - dep[j] - 1 == d) {
            node = j;
            break;
          }
          if (dep[j] + 1 < m) m = dep[j] + 1, node = j;
        }
      --deg[par[i]];
      ++deg[node];
      d -= dep[i] - dep[node] - 1;
      dep[i] = dep[node] + 1;
      par[i] = node;
    }
    for (int i = 1; i < n; ++i) cout << par[i] + 1 << (i != n - 1 ? ' ' : '\n');
  }
}

### Prompt

In cpp, your task is to solve 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() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  for (; t--;) {
    int n, d;
    cin >> n >> d;
    int Min = 0, Max = n * (n - 1) / 2, now = 0;
    for (int i = 2; i <= n; ++i) {
      if ((1 << (now + 1)) == i) ++now;
      Min += now;
    }
    if (d < Min || Max < d) {
      cout << "NO" << '\n';
      continue;
    }
    cout << "YES" << '\n';
    d = Max - d;
    vector<int> par(n), dep(n), deg(n, 1);
    iota(par.begin(), par.end(), -1);
    iota(dep.begin(), dep.end(), 0);
    deg[n - 1] = 0;
    for (int i = n - 1; i > 0; --i) {
      if (!d) break;
      int m = n, node;
      for (int j = 0; j < n; ++j)
        if (j != i && deg[j] != 2) {
          if (dep[i] - dep[j] - 1 == d) {
            node = j;
            break;
          }
          if (dep[j] + 1 < m) m = dep[j] + 1, node = j;
        }
      --deg[par[i]];
      ++deg[node];
      d -= dep[i] - dep[node] - 1;
      dep[i] = dep[node] + 1;
      par[i] = node;
    }
    for (int i = 1; i < n; ++i) cout << par[i] + 1 << (i != n - 1 ? ' ' : '\n');
  }
}

```