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>
#include <bits/stdc++.h>
using namespace std;
int t, n, d;
int p[5010], dep[5010], deg[5010];
bitset<5010> no;
int getLow(int x) {
  int ret = 0, k = 0;
  while (x) {
    ret += min(x, (1 << k)) * k;
    x = max(0, x - (1 << k));
    ++k;
  }
  return ret;
}
void solve() {
  cin >> n >> d;
  if (d < getLow(n) || n * (n - 1) / 2 < d) {
    cout << "NO\n";
    return;
  }
  for (int i = 1; i <= n; i++) no[i] = false;
  deg[1] = 2;
  for (int i = 2; i <= n; i++) p[i] = i - 1, dep[i] = i - 1, deg[i] = 2;
  deg[n] = 1;
  int cnt = n * (n - 1) / 2;
  dep[0] = 1e9;
  while (cnt > d) {
    int v = 0;
    for (int i = 1; i <= n; i++) {
      if (!no[i] && deg[i] == 1 && dep[v] > dep[i]) v = i;
    }
    if (v == 0) {
      cout << "NO\n";
      return;
    }
    for (int i = 1; i <= n; i++) {
      if (deg[i] <= 2 && dep[i] == dep[v] - 2) {
        deg[p[v]]--;
        deg[i]++;
        dep[v]--;
        p[v] = i;
        break;
      }
      if (i == n) no[v] = true;
    }
    if (!no[v]) --cnt;
  }
  cout << "YES\n";
  for (int i = 2; i <= n; i++) cout << p[i] << " \n"[i == n];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> t;
  while (t--) solve();
  return 0;
}
