#include <bits/stdc++.h>
using namespace std;
void testCase() {
  int n, d;
  cin >> n >> d;
  int total = n * (n - 1) / 2;
  if (d > total) {
    cout << "NO\n";
    return;
  }
  vector<int> v;
  for (int i = 0; i < n; i++) v.push_back(1);
  int taso = 0;
  int kohta = n - 1;
  while (total > d) {
    if (v[taso] >= (1 << taso)) taso++;
    if (taso >= kohta) {
      cout << "NO\n";
      return;
    }
    int ero = kohta - taso;
    if (total - d > ero) {
      total -= ero;
      v[kohta]--;
      v[taso]++;
    } else {
      int monta = total - d;
      v[kohta]--;
      v[kohta - monta]++;
      total = d;
      break;
    }
    kohta--;
  }
  cout << "YES\n";
  vector<int> prevtaso;
  prevtaso.push_back(1);
  int solmu = 2;
  for (int i = 1; i < v.size(); i++) {
    if (v[i] == 0) break;
    vector<int> uustaso;
    int ind = -1;
    for (int j = 0; j < v[i]; j++) {
      if (j % 2 == 0) ind++;
      cout << prevtaso[ind] << " ";
      uustaso.push_back(solmu);
      solmu++;
    }
    prevtaso = uustaso;
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) {
    testCase();
  }
}

### Prompt

Please create a solution in Cpp 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;
void testCase() {
  int n, d;
  cin >> n >> d;
  int total = n * (n - 1) / 2;
  if (d > total) {
    cout << "NO\n";
    return;
  }
  vector<int> v;
  for (int i = 0; i < n; i++) v.push_back(1);
  int taso = 0;
  int kohta = n - 1;
  while (total > d) {
    if (v[taso] >= (1 << taso)) taso++;
    if (taso >= kohta) {
      cout << "NO\n";
      return;
    }
    int ero = kohta - taso;
    if (total - d > ero) {
      total -= ero;
      v[kohta]--;
      v[taso]++;
    } else {
      int monta = total - d;
      v[kohta]--;
      v[kohta - monta]++;
      total = d;
      break;
    }
    kohta--;
  }
  cout << "YES\n";
  vector<int> prevtaso;
  prevtaso.push_back(1);
  int solmu = 2;
  for (int i = 1; i < v.size(); i++) {
    if (v[i] == 0) break;
    vector<int> uustaso;
    int ind = -1;
    for (int j = 0; j < v[i]; j++) {
      if (j % 2 == 0) ind++;
      cout << prevtaso[ind] << " ";
      uustaso.push_back(solmu);
      solmu++;
    }
    prevtaso = uustaso;
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) {
    testCase();
  }
}

```