#include <bits/stdc++.h>
using namespace std;
const int N = 5005;
int dep[N], p[N];
int n, d;
vector<int> v_in_dep[N];
int main() {
  ios::sync_with_stdio(0);
  int T;
  cin >> T;
  while (T--) {
    cin >> n >> d;
    for (int i = 0; i <= n; i++) {
      v_in_dep[i].clear();
    }
    dep[0] = -1;
    int now = 0;
    if (n * (n - 1) / 2 < d) {
      cout << "NO" << endl;
      continue;
    }
    for (int i = 1; i <= n; i++) {
      p[i] = i / 2;
      dep[i] = dep[p[i]] + 1;
      v_in_dep[dep[i]].push_back(i);
      now += dep[i];
    }
    if (d < now) {
      cout << "NO" << endl;
      continue;
    }
    cout << "YES" << endl;
    d -= now;
    while (d) {
      int q = n;
      while (v_in_dep[q].size() < 2) q--;
      if (v_in_dep[q].size() == 1) {
        q--;
        continue;
      }
      int j = v_in_dep[q].back();
      v_in_dep[q].pop_back();
      p[j] = v_in_dep[q][0];
      v_in_dep[q + 1].push_back(j);
      d--;
    }
    for (int i = 2; i <= n; i++) cout << p[i] << " ";
    cout << endl;
  }
  return 0;
}

### 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;
const int N = 5005;
int dep[N], p[N];
int n, d;
vector<int> v_in_dep[N];
int main() {
  ios::sync_with_stdio(0);
  int T;
  cin >> T;
  while (T--) {
    cin >> n >> d;
    for (int i = 0; i <= n; i++) {
      v_in_dep[i].clear();
    }
    dep[0] = -1;
    int now = 0;
    if (n * (n - 1) / 2 < d) {
      cout << "NO" << endl;
      continue;
    }
    for (int i = 1; i <= n; i++) {
      p[i] = i / 2;
      dep[i] = dep[p[i]] + 1;
      v_in_dep[dep[i]].push_back(i);
      now += dep[i];
    }
    if (d < now) {
      cout << "NO" << endl;
      continue;
    }
    cout << "YES" << endl;
    d -= now;
    while (d) {
      int q = n;
      while (v_in_dep[q].size() < 2) q--;
      if (v_in_dep[q].size() == 1) {
        q--;
        continue;
      }
      int j = v_in_dep[q].back();
      v_in_dep[q].pop_back();
      p[j] = v_in_dep[q][0];
      v_in_dep[q + 1].push_back(j);
      d--;
    }
    for (int i = 2; i <= n; i++) cout << p[i] << " ";
    cout << endl;
  }
  return 0;
}

```