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 ans[5005];
int vis[5005];
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, d;
    cin >> n >> d;
    for (int i = 2; i <= n; i++) ans[i] = i - 1, vis[i] = 0;
    vis[1] = 0;
    int now = n * (n - 1) / 2;
    if (now < d) {
      printf("NO\n");
      continue;
    }
    int up = 1, dn = n;
    bool f = true;
    queue<int> q, qq;
    q.push(1);
    while (now != d) {
      if (up + 1 == dn) {
        f = false;
        break;
      }
      int res = dn - up - 1;
      if (res > now - d) {
        res = now - d;
        ans[dn] = dn - res - 1;
        break;
      }
      if (q.empty()) {
        while (!qq.empty()) q.push(qq.front()), qq.pop();
        up++;
        q.push(up);
        continue;
      }
      ans[dn] = q.front();
      q.pop();
      qq.push(dn);
      qq.push(dn);
      dn--;
      now -= res;
    }
    if (f) {
      printf("YES\n");
      for (int i = 2; i <= n; i++) printf("%d ", ans[i]);
      printf("\n");
    } else
      printf("NO\n");
  }
}