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>
#include <bits/stdc++.h>
const double eps = 1e-10;
const int maxn = 5e3 + 10;
;
const long long mod = 1e9 + 7;
int sgn(double a) { return a < -eps ? -1 : a < eps ? 0 : 1; }
using namespace std;
int fa[maxn], dep[maxn];
vector<int> vec[maxn];
int main() {
  int T;
  cin >> T;
  while (T--) {
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) dep[i] = 0, fa[i] = 0, vec[i].clear();
    vec[0].push_back(1);
    int maxx = 0;
    for (int i = 2; i <= n; i++) {
      dep[i] = dep[i >> 1] + 1;
      vec[dep[i]].push_back(i);
      maxx = max(maxx, dep[i]);
      m -= dep[i];
    }
    if (m < 0) {
      cout << "NO" << endl;
      continue;
    }
    bool flag = true;
    while (m) {
      int change = maxx;
      while (change >= 0 && vec[change].size() == 1) change--;
      if (change <= 0) {
        cout << "NO" << endl;
        flag = false;
        break;
      }
      int nod = vec[change][vec[change].size() - 1];
      vec[change].pop_back();
      vec[change + 1].push_back(nod);
      maxx = max(maxx, change + 1);
      m--;
    }
    if (flag) {
      cout << "YES" << endl;
      for (int i = 1; i <= n; i++) {
        for (int j = 0; j < vec[i].size(); j++)
          fa[vec[i][j]] = vec[i - 1][j >> 1];
      }
      for (int i = 2; i <= n; i++) {
        cout << fa[i] << " ";
      }
      cout << endl;
    }
  }
  return 0;
}
