#include <bits/stdc++.h>
using namespace std;
vector<int> depth[5005];
int p[5005];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int ntest;
  cin >> ntest;
  while (ntest--) {
    int n, d;
    cin >> n >> d;
    if (d > n * (n - 1) / 2) {
      cout << "NO\n";
      continue;
    }
    for (int i = 0; i <= n; i++) depth[i].clear();
    depth[0].push_back(1);
    int cur = 1;
    for (int i = 2; i <= n; i++) {
      depth[cur].push_back(i);
      if (((int)(depth[cur]).size()) == 2 * ((int)(depth[cur - 1]).size()))
        cur++;
    }
    if (((int)(depth[cur]).size()) == 0) cur--;
    int curD = 0;
    for (int i = 1; i <= cur; i++) {
      curD += ((int)(depth[i]).size()) * i;
    }
    if (curD > d) {
      cout << "NO\n";
      continue;
    }
    cout << "YES\n";
    while (curD < d) {
      for (int i = cur; i >= 1; i--) {
        if (depth[i].size() != 1) {
          int u = depth[i].back();
          depth[i].pop_back();
          depth[i + 1].push_back(u);
          break;
        }
      }
      if (depth[cur + 1].size()) cur++;
      curD++;
    }
    for (int i = cur; i >= 1; i--) {
      for (int j = 0; j < ((int)(depth[i]).size()); j++) {
        p[depth[i][j]] = depth[i - 1][j / 2];
      }
    }
    for (int i = 2; i <= n; i++) {
      cout << p[i] << ' ';
    }
    cout << '\n';
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
vector<int> depth[5005];
int p[5005];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int ntest;
  cin >> ntest;
  while (ntest--) {
    int n, d;
    cin >> n >> d;
    if (d > n * (n - 1) / 2) {
      cout << "NO\n";
      continue;
    }
    for (int i = 0; i <= n; i++) depth[i].clear();
    depth[0].push_back(1);
    int cur = 1;
    for (int i = 2; i <= n; i++) {
      depth[cur].push_back(i);
      if (((int)(depth[cur]).size()) == 2 * ((int)(depth[cur - 1]).size()))
        cur++;
    }
    if (((int)(depth[cur]).size()) == 0) cur--;
    int curD = 0;
    for (int i = 1; i <= cur; i++) {
      curD += ((int)(depth[i]).size()) * i;
    }
    if (curD > d) {
      cout << "NO\n";
      continue;
    }
    cout << "YES\n";
    while (curD < d) {
      for (int i = cur; i >= 1; i--) {
        if (depth[i].size() != 1) {
          int u = depth[i].back();
          depth[i].pop_back();
          depth[i + 1].push_back(u);
          break;
        }
      }
      if (depth[cur + 1].size()) cur++;
      curD++;
    }
    for (int i = cur; i >= 1; i--) {
      for (int j = 0; j < ((int)(depth[i]).size()); j++) {
        p[depth[i][j]] = depth[i - 1][j / 2];
      }
    }
    for (int i = 2; i <= n; i++) {
      cout << p[i] << ' ';
    }
    cout << '\n';
  }
  return 0;
}

```