#include <bits/stdc++.h>
using namespace std;
int dep[5555];
int son[5555], fa[5444];
int bad[5555];
int main() {
  ios::sync_with_stdio(0);
  long long t;
  cin >> t;
  dep[0] = 0x6fffffff;
  while (t--) {
    int n, d;
    cin >> n >> d;
    int pd = 0;
    for (int i = 1, cnt = 0; i <= n; i++) {
      if (!(i & (i - 1))) cnt++;
      pd += cnt - 1;
    }
    if (d < pd || d > (n * (n - 1) / 2)) {
      cout << "NO" << endl;
      continue;
    }
    for (int i = 1; i <= n; i++) {
      dep[i] = i - 1;
      son[i - 1] = 1;
      fa[i] = i - 1;
      bad[i] = 0;
    }
    son[n] = 0;
    int t = (n - 1) * n / 2;
    while (t > d) {
      int c = 0;
      for (int i = 1; i <= n; i++) {
        if (son[i] == 0 && !bad[i] && dep[i] < dep[c]) {
          c = i;
        }
      }
      assert(c);
      int p = 0;
      for (int i = 1; i <= n; i++) {
        if (dep[i] + 2 == dep[c] && son[i] < 2) p = i;
      }
      if (!p || dep[c] <= 1) {
        bad[c] = 1;
        continue;
      }
      son[fa[c]]--;
      fa[c] = p;
      son[p]++;
      dep[c]--;
      t--;
    }
    cout << "YES" << endl;
    for (int i = 2; i <= n; i++) cout << fa[i] << " ";
    cout << endl;
  }
  return 0;
}

### 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;
int dep[5555];
int son[5555], fa[5444];
int bad[5555];
int main() {
  ios::sync_with_stdio(0);
  long long t;
  cin >> t;
  dep[0] = 0x6fffffff;
  while (t--) {
    int n, d;
    cin >> n >> d;
    int pd = 0;
    for (int i = 1, cnt = 0; i <= n; i++) {
      if (!(i & (i - 1))) cnt++;
      pd += cnt - 1;
    }
    if (d < pd || d > (n * (n - 1) / 2)) {
      cout << "NO" << endl;
      continue;
    }
    for (int i = 1; i <= n; i++) {
      dep[i] = i - 1;
      son[i - 1] = 1;
      fa[i] = i - 1;
      bad[i] = 0;
    }
    son[n] = 0;
    int t = (n - 1) * n / 2;
    while (t > d) {
      int c = 0;
      for (int i = 1; i <= n; i++) {
        if (son[i] == 0 && !bad[i] && dep[i] < dep[c]) {
          c = i;
        }
      }
      assert(c);
      int p = 0;
      for (int i = 1; i <= n; i++) {
        if (dep[i] + 2 == dep[c] && son[i] < 2) p = i;
      }
      if (!p || dep[c] <= 1) {
        bad[c] = 1;
        continue;
      }
      son[fa[c]]--;
      fa[c] = p;
      son[p]++;
      dep[c]--;
      t--;
    }
    cout << "YES" << endl;
    for (int i = 2; i <= n; i++) cout << fa[i] << " ";
    cout << endl;
  }
  return 0;
}

```