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 = 5000 + 10;
int dep[N], vis[N];
void solve() {
  int n, d;
  cin >> n >> d;
  if (d > n * (n - 1) / 2) return cout << "NO\n", void();
  int now = 0, cnt = 1;
  for (int i = 1; i <= n; i++) {
    if (cnt == 0) now++, cnt = (1 << now);
    dep[i] = now, cnt--;
  }
  for (int i = 1; i <= n; i++) d -= dep[i];
  if (d < 0) return cout << "NO\n", void();
  for (int i = 1; i <= n; i++) vis[i] = 0;
  int mxdep = dep[n];
  for (int i = n; i >= 2; i--) {
    if (dep[i] == dep[i - 1]) {
      if (d > mxdep + 1 - dep[i]) {
        d -= mxdep + 1 - dep[i];
        mxdep++;
        dep[i] = mxdep;
      } else {
        dep[i] += d;
        d = 0;
        break;
      }
    }
  }
  sort(dep + 1, dep + n + 1);
  now = 1;
  cout << "YES\n";
  for (int i = 2; i <= n; i++) {
    while (vis[now] >= 2 || dep[now] + 1 < dep[i]) now++;
    cout << now << " \n"[i == n], vis[now]++;
  }
  return;
}
int main() {
  int T;
  for (scanf("%d", &T); T--;) {
    solve();
  }
  return 0;
}