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:

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#include <bits/stdc++.h>
using namespace std;
long long dp[5001];
pair<long long, vector<pair<long long, long long>>> solve(long long n,
                                                          long long d) {
  if (d < dp[n] || d > n * (n - 1) / 2) assert(0);
  if (n == 0) return {-1, {}};
  if (n == 1) return {0, {}};
  if (n == 2) return {0, {{0, 1}}};
  d -= n - 1;
  for (long long i = 0; i < n; i++) {
    long long j = n - i - 1;
    long long mn = dp[i] + dp[j];
    long long mx = i * (i - 1) / 2 + j * (j - 1) / 2;
    if (mn <= d && d <= mx) {
      long long d1 = dp[i], d2 = dp[j];
      long long first = min(i * (i - 1) / 2 - d1, d - d1 - d2);
      d1 += first;
      long long second = min(j * (j - 1) / 2 - d2, d - d1 - d2);
      d2 += second;
      auto l = solve(i, d1), r = solve(j, d2);
      if (r.first != -1) r.first += i + 1;
      for (auto &el : r.second) {
        el.first += i + 1;
        el.second += i + 1;
      }
      if (l.first != -1) l.first++;
      for (auto &el : l.second) {
        el.first++;
        el.second++;
      }
      pair<long long, vector<pair<long long, long long>>> ans = l;
      ans.first = 0;
      if (l.first != -1) ans.second.push_back({l.first, 0});
      if (r.first != -1) ans.second.push_back({r.first, 0});
      for (auto el : r.second) ans.second.push_back(el);
      return ans;
    }
  }
}
long long pr[5000], h[5000];
vector<long long> g[5000];
void dfs(long long v) {
  for (auto u : g[v]) {
    if (u != pr[v]) {
      pr[u] = v;
      h[u] = h[v] + 1;
      dfs(u);
    }
  }
}
void solve() {
  long long n, d;
  cin >> n >> d;
  if (d < dp[n] || d > n * (n - 1) / 2)
    cout << "NO\n";
  else {
    for (long long i = 0; i < n; i++) g[i].clear();
    cout << "YES\n";
    auto ans = solve(n, d);
    pr[0] = -1;
    for (auto el : ans.second) {
      g[el.first].push_back(el.second);
      g[el.second].push_back(el.first);
    }
    h[0] = 0;
    dfs(0);
    for (long long i = 1; i < n; i++) cout << pr[i] + 1 << " ";
    cout << '\n';
    long long sumh = 0;
    for (long long i = 0; i < n; i++) sumh += h[i];
    assert(sumh == d);
  }
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  dp[0] = 0;
  dp[1] = 0;
  dp[2] = 1;
  for (long long i = 3; i <= 5000; i++) {
    dp[i] = 5001;
    for (long long j = 0; j < i; j++)
      dp[i] = min(dp[i], dp[j] + dp[i - j - 1] + i - 1);
  }
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}
