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;
using ll = long long;
using ii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vll = vector<ll>;
const ll mod = 7 + 1e9;
const int inf = 1e9 + 5;
const ll INF = ll(1e18) + 5;
const int N = 2e5 + 5;
void solve() {
  int n, d;
  cin >> n >> d;
  int mx = n * (n - 1) / 2;
  int mn = 0, cnt = 0, cur = 0;
  for (int i = 0; i < n; ++i) {
    mn += cnt;
    cur++;
    if (cur == (1 << cnt)) {
      cnt++;
      cur = 0;
    }
  }
  if (d > mx || d < mn) {
    cout << "NO\n";
    return;
  }
  vector<int> par(n, 0), depth(n);
  iota(par.begin() + 1, par.end(), 0);
  iota(depth.begin(), depth.end(), 0);
  vector<int> deg(n, 1);
  deg[n - 1] = 0;
  vector<bool> good(n, 1);
  while (mx > d) {
    int u = -1;
    for (int i = 0; i < n; ++i) {
      if (good[i] && deg[i] == 0 && (u == -1 || depth[u] > depth[i])) {
        u = i;
      }
    }
    assert(u != -1);
    int to = -1, mx_d = 0;
    for (int i = 0; i < n; ++i) {
      if (deg[i] < 2 && depth[u] - depth[i] - 1 <= mx - d &&
          depth[u] - depth[i] - 1 > mx_d && depth[u] - depth[i] >= 2 &&
          i != u) {
        to = i;
        mx_d = depth[u] - depth[i] - 1;
      }
    }
    if (to == -1) {
      good[u] = 0;
      continue;
    }
    mx -= mx_d;
    deg[par[u]]--;
    par[u] = to;
    deg[to]++;
    depth[u] -= mx_d;
  }
  cout << "YES\n";
  for (int i = 1; i < n; ++i) {
    cout << par[i] + 1 << " \n"[i == n - 1];
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}