#include <bits/stdc++.h>
using namespace std;
vector<long long> graph[5005];
long long par[5005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long l, r, i, j, n, m, t, d, a;
  cin >> t;
  for (long long cz = 0; cz < t; cz++) {
    cin >> n >> d;
    long long ans = 0;
    if ((n * (n - 1)) / 2 < d) {
      cout << "NO" << endl;
      continue;
    }
    i = 1;
    m = n;
    m--;
    while (m) {
      if (m >= (1 << i)) {
        ans += (i) * (1 << i);
        m -= (1 << i);
      } else {
        ans += m * (i);
        m = 0;
      }
      i++;
    }
    if (ans > d) {
      cout << "NO" << endl;
      continue;
    }
    ans = 0;
    cout << "YES" << endl;
    for (i = 1; i <= n; i++) graph[i].clear();
    for (i = 1; i <= n; i++) {
      graph[i].push_back(i);
      ans += (i - 1);
    }
    long long l = 2;
    for (i = n; i >= 1; i--) {
      if (ans - (i - l) >= d) {
        a = graph[i][graph[i].size() - 1];
        graph[i].pop_back();
        graph[l].push_back(a);
        ans -= (i - l);
        if (graph[l].size() == (1 << (l - 1))) l++;
        if (ans == d) break;
      } else {
        a = graph[i][graph[i].size() - 1];
        graph[i].pop_back();
        graph[i - (ans - d)].push_back(a);
        break;
      }
    }
    for (i = 2; i <= n; i++) {
      for (j = 0; j < graph[i].size(); j++) {
        par[graph[i][j]] = graph[i - 1][j / 2];
      }
    }
    for (i = 2; i <= n; i++) cout << par[i] << " ";
    cout << endl;
  }
}

### Prompt

Develop a solution in cpp 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<long long> graph[5005];
long long par[5005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long l, r, i, j, n, m, t, d, a;
  cin >> t;
  for (long long cz = 0; cz < t; cz++) {
    cin >> n >> d;
    long long ans = 0;
    if ((n * (n - 1)) / 2 < d) {
      cout << "NO" << endl;
      continue;
    }
    i = 1;
    m = n;
    m--;
    while (m) {
      if (m >= (1 << i)) {
        ans += (i) * (1 << i);
        m -= (1 << i);
      } else {
        ans += m * (i);
        m = 0;
      }
      i++;
    }
    if (ans > d) {
      cout << "NO" << endl;
      continue;
    }
    ans = 0;
    cout << "YES" << endl;
    for (i = 1; i <= n; i++) graph[i].clear();
    for (i = 1; i <= n; i++) {
      graph[i].push_back(i);
      ans += (i - 1);
    }
    long long l = 2;
    for (i = n; i >= 1; i--) {
      if (ans - (i - l) >= d) {
        a = graph[i][graph[i].size() - 1];
        graph[i].pop_back();
        graph[l].push_back(a);
        ans -= (i - l);
        if (graph[l].size() == (1 << (l - 1))) l++;
        if (ans == d) break;
      } else {
        a = graph[i][graph[i].size() - 1];
        graph[i].pop_back();
        graph[i - (ans - d)].push_back(a);
        break;
      }
    }
    for (i = 2; i <= n; i++) {
      for (j = 0; j < graph[i].size(); j++) {
        par[graph[i][j]] = graph[i - 1][j / 2];
      }
    }
    for (i = 2; i <= n; i++) cout << par[i] << " ";
    cout << endl;
  }
}

```