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>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    long long int i, j, n, d, m;
    cin >> n >> d;
    j = (n * (n - 1));
    j /= 2;
    long long int count1 = 1;
    vector<long long int> ar(n - 1);
    for (i = 0; i < n - 1; i++) {
      ar[i] = i + 1;
    }
    vector<list<long long int> > arr(n);
    for (i = 0; i < n - 1; i++) {
      arr[i].push_back(i + 2);
    }
    vector<list<long long int> > ar1(n);
    for (i = 0; i < n; i++) {
      ar1[i].push_back(i + 1);
    }
    vector<long long int> ar2(n);
    for (i = 0; i < n; i++) {
      ar2[i] = i;
    }
    vector<long long int> his(n);
    for (i = 0; i < n; i++) {
      his[i] = 1;
    }
    long long int x = n;
    while (j != d && x != 1) {
      long long int y = ar2[x - 1];
      if (y < 2) {
        x--;
        continue;
      } else {
        long long int z = y - 2;
        long long int count1 = 0;
        long long int z1;
        while (z >= 0) {
          for (auto it = ar1[z].begin(); it != ar1[z].end(); it++) {
            z1 = *it;
            if (arr[z1 - 1].size() >= 2) {
              continue;
            } else {
              count1 = 1;
              break;
            }
          }
          if (count1 == 1) {
            break;
          } else {
            z--;
          }
        }
        if (count1 == 1) {
          arr[z1 - 1].push_back(x);
          arr[ar[x - 2] - 1].remove(x);
          ar[x - 2] = z1;
          ar1[y].remove(x);
          ar2[x - 1] = z + 1;
          ar1[z + 1].push_back(x);
          j--;
        } else {
          x--;
          continue;
        }
      }
    }
    if (j == d) {
      cout << "YES\n";
      for (i = 0; i < n - 1; i++) {
        cout << ar[i] << " ";
      }
      cout << "\n";
    } else {
      cout << "NO\n";
    }
  }
}