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;
const int N = 1e5 + 5;
vector<int> g[N];
int parent[N];
bool pos(int n, int d) {
  int sum = n * (n - 1) / 2;
  if (sum < d) return 0;
  int x = sum - d;
  int last = 1, now, mx = n - 1;
  g[0].push_back(1);
  for (int i = 1; i <= mx; i++) {
    g[i].push_back(i + 1);
    now = 1;
    while (now < last * 2 && mx - i <= x && mx > i) {
      now++;
      x -= mx - i;
      g[i].push_back(mx + 1);
      mx--;
    }
    last = now;
  }
  if (x) return 0;
  return 1;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, d;
    cin >> n >> d;
    if (pos(n, d)) {
      cout << "YES" << endl;
      for (int i = 1; i <= n; i++) {
        for (int j = 0; j < g[i].size(); j++) parent[g[i][j]] = g[i - 1][j / 2];
      }
      for (int i = 2; i <= n; i++) cout << parent[i] << " ";
      cout << endl;
    } else
      cout << "NO" << endl;
    for (int i = 0; i <= n; i++) g[i].clear();
  }
  return 0;
}