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 int inf = 1e9 + 7;
const long long longinf = 1LL << 60;
const long long mod = 1e9 + 7;
const long double eps = 1e-10;
template <typename T1, typename T2>
inline void chmin(T1 &a, T2 b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
inline void chmax(T1 &a, T2 b) {
  if (a < b) a = b;
}
void solve() {
  int n, d;
  cin >> n >> d;
  vector<int> cnt(n, 1);
  int l = 1, r = n - 1;
  long long sum = n * (n - 1) / 2;
  while (sum > d) {
    if (l >= r) break;
    if (sum - d > r - l) {
      sum -= r - l;
      cnt[r]--;
      cnt[l]++;
      r--;
      if (cnt[l] == (long long)pow(2, l)) l++;
    } else {
      cnt[r]--;
      cnt[r - (sum - d)]++;
      sum = d;
    }
  }
  if (sum == d) {
    cout << "YES" << '\n';
    assert(accumulate(cnt.begin(), cnt.end(), 0) == n);
    vector<int> ans(n);
    vector<vector<int>> mem(n);
    int id = 0;
    for (int i = (int)(0); i < (int)(n); i++)
      for (int j = (int)(0); j < (int)(cnt[i]); j++) mem[i].push_back(id++);
    for (int i = (int)(0); i < (int)(n); i++)
      if (i) {
        int j = 0;
        int cnt = 0;
        for (int id : mem[i]) {
          ans[id] = mem[i - 1][j];
          cnt++;
          if (cnt == 2) {
            j++;
            cnt = 0;
          }
        }
      }
    for (int i = (int)(0); i < (int)(n); i++)
      if (i) cout << ans[i] + 1 << ' ';
    cout << '\n';
  } else {
    cout << "NO" << '\n';
  }
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int T;
  cin >> T;
  while (T--) solve();
  return 0;
}