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;
mt19937_64 mrand(chrono::steady_clock::now().time_since_epoch().count());
int fa[5005], dep[5005];
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, d;
    cin >> n >> d;
    int max_dep = n * (n - 1) / 2;
    if (d > max_dep) {
      cout << "NO\n";
      continue;
    }
    int mx = 0;
    for (int i = 2; i <= n; ++i)
      fa[i] = i / 2, dep[i] = dep[fa[i]] + 1, mx = max(mx, dep[i]);
    int min_dep = 0;
    for (int i = 1; i <= n; ++i) min_dep += dep[i];
    if (d < min_dep) {
      cout << "NO\n";
      continue;
    }
    cout << "YES\n";
    int cur = min_dep, need = d - cur, lst = 1 << (31 - __builtin_clz(n));
    for (int i = n; i >= 1; --i) {
      if (!need) break;
      if (__builtin_popcount(i) == 1) continue;
      int f = fa[i], df = dep[f], dd = dep[lst] - dep[f];
      if (need <= dd) {
        int tp = dd - need;
        int ff = lst;
        while (tp--) ff = fa[ff];
        fa[i] = ff;
        dep[i] = dep[ff] + 1;
        break;
      } else {
        fa[i] = lst;
        dep[i] = dep[lst] + 1;
        need -= dd;
      }
      lst = i;
    }
    int sum = 0;
    for (int i = 2; i <= n; ++i) {
      cout << fa[i] << " ";
      assert(dep[i] == dep[fa[i]] + 1);
      sum += dep[i];
    }
    cout << endl;
    assert(sum == d);
  }
}