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 fa[5010], p[5010];
vector<int> v[5010];
int main() {
  int T;
  cin >> T;
  while (T--) {
    int n, d;
    cin >> n >> d;
    memset(fa, 0, sizeof(fa));
    memset(p, 0, sizeof(p));
    int sum = 0, y = 0;
    for (int i = 1; i <= n;) {
      int x = min(n - i + 1, 1 << y);
      p[y] = x, sum += x * y;
      i += x, y++;
    }
    if (sum > d) {
      cout << "NO" << endl;
      continue;
    }
    while (sum < d) {
      int i = 0;
      for (int k = 1; k <= n; k++)
        if ((p[k] - 1) * 2 >= p[k + 1] + 1) {
          i = k;
          break;
        }
      if (i == 0) break;
      sum++, p[i]--, p[i + 1]++;
    }
    if (sum < d) {
      cout << "NO" << endl;
      continue;
    }
    cout << "YES" << endl;
    int cnt = 1;
    for (int i = 0; i <= n; i++) v[i].clear();
    v[0].push_back(cnt);
    v[0].push_back(cnt);
    for (int i = 1; i <= n; i++) {
      while (p[i]--) {
        fa[++cnt] = v[i - 1].back();
        v[i - 1].pop_back();
        v[i].push_back(cnt);
        v[i].push_back(cnt);
      }
    }
    for (int i = 2; i <= n; i++) cout << fa[i] << " ";
    cout << "\n";
  }
  return 0;
}