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;
int main() {
  std::ios_base::sync_with_stdio(false);
  int T;
  cin >> T;
  vector<int> f(5001), g(5001);
  for (int i = 1; i <= 5000; i++) {
    f[i] = g[i] = (i - 1) * i / 2;
    for (int j = 0; j < i; j++) {
      f[i] = min(f[i], f[j] + f[i - j - 1] + i - 1);
    }
  }
  while (T--) {
    int n, D;
    cin >> n >> D;
    if (D < f[n] || D > g[n]) {
      cout << "NO" << endl;
      continue;
    }
    vector<int> pre(n + 1);
    function<void(int, int, int, int)> solve = [&](int l, int r, int fa,
                                                   int sum) {
      if (l > r) return;
      pre[l] = fa;
      int z = r - l + 1;
      assert(f[z] <= sum && g[z] >= sum);
      sum -= (z - 1);
      for (int i = 0; i < z; i++) {
        if (f[z - i - 1] <= sum - f[i] && g[z - i - 1] >= sum - f[i]) {
          solve(l + 1, l + i, l, f[i]);
          solve(l + i + 1, r, l, sum - f[i]);
          break;
        } else if (f[z - i - 1] <= sum - g[i] && g[z - i - 1] >= sum - g[i]) {
          solve(l + 1, l + i, l, g[i]);
          solve(l + i + 1, r, l, sum - g[i]);
          break;
        }
      }
    };
    cout << "YES" << endl;
    solve(1, n, 0, D);
    for (int i = 2; i <= n; i++) {
      assert(pre[i]);
      cout << pre[i] << " ";
    }
    cout << endl;
  }
  return 0;
}