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;
template <typename T>
void out(T x) {
  cout << x << endl;
  exit(0);
}
const int maxn = 1e6 + 5;
bool pw2(int i) { return (i & (i - 1)) == 0; }
int n, d;
void solve() {
  cin >> n >> d;
  int cur = n * (n - 1) / 2;
  if (d > cur) {
    cout << "NO\n";
    return;
  }
  int lb = 0;
  for (int i = 1, r = 0; i <= n; i++) {
    if (pw2(i)) r++;
    lb += r - 1;
  }
  if (d < lb) {
    cout << "NO\n";
    return;
  }
  vector<int> par(n), ch(n), depth(n);
  vector<bool> done(n, false);
  for (int i = 0; i < n; i++) {
    par[i] = i - 1;
    depth[i] = i;
    ch[i] = i < n - 1 ? 1 : 0;
  }
  while (cur > d) {
    int x = -1;
    for (int i = 0; i < n; i++) {
      if (ch[i] == 0 && !done[i] && (x == -1 || depth[x] > depth[i])) {
        x = i;
      }
    }
    if (x == -1) break;
    int p = -1;
    for (int i = 0; i < n; i++) {
      if (depth[i] <= depth[x] - 2 && ch[i] < 2 &&
          (p == -1 || depth[p] < depth[i])) {
        p = i;
      }
    }
    if (p == -1) {
      done[x] = true;
      continue;
    }
    assert(x != -1 && p != -1);
    cur--;
    ch[par[x]]--;
    par[x] = p;
    ch[p]++;
    depth[x]--;
  }
  if (cur > d) {
    cout << "NO\n";
    return;
  }
  cout << "YES\n";
  for (int i = 1; i < n; i++) {
    cout << par[i] + 1 << " ";
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}