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 long double EPS = 1e-14;
const int maxn = 100500;
const int mod = 1e9 + 7;
bool bad[5010];
int p[5010], dep[5010], child[5010];
set<pair<int, int> > dp[5010];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n, d;
    cin >> n >> d;
    for (int i = 0; i < n; ++i)
      dep[i] = 0, child[i] = 0, bad[i] = false, dp[i].clear();
    bad[0] = true;
    for (int i = 1; i < n; ++i) {
      p[i] = i - 1;
      d -= i;
      dep[i] = i;
      child[i - 1] = 1;
      dp[i - 1].insert({1, i - 1});
    }
    dp[n - 1].insert({0, n - 1});
    set<pair<int, int> > lv;
    lv.insert({n - 1, n - 1});
    if (d > 0) {
      cout << "NO\n";
      continue;
    }
    while (d < 0 && (int)(lv).size()) {
      auto u = *lv.begin();
      lv.erase(u);
      int lst = -1;
      int idx = u.first - 2;
      while (u.first - idx - 1 <= -d && idx >= 0 && (int)(dp[idx]).size() &&
             dp[idx].begin()->first < 2) {
        lst = idx;
        idx--;
      }
      if (~lst) {
        d += u.first - lst - 1;
        auto v = *dp[lst].begin();
        dp[lst].erase(v);
        v.first++;
        if (v.first < 2) dp[lst].insert(v);
        dp[u.first].erase({0, u.second});
        dp[lst + 1].insert({0, u.second});
        int i = v.second;
        dp[u.first - 1].erase({child[p[u.second]], p[u.second]});
        child[p[u.second]]--;
        dp[u.first - 1].insert({child[p[u.second]], p[u.second]});
        if (child[p[u.second]] == 0 && !bad[p[u.second]])
          lv.insert({dep[p[u.second]], p[u.second]});
        child[i]++;
        if (child[i] == 1) lv.erase({dep[i], i});
        p[u.second] = i;
        dep[u.second] = dep[i] + 1;
      }
    }
    if (!d) {
      cout << "YES\n";
      for (int i = 1; i < n; ++i) cout << p[i] + 1 << ' ';
      cout << '\n';
    } else
      cout << "NO\n";
  }
  return 0;
}