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 MAXN = 200009;
const long long MOD = 119 << 23 | 1;
class {
 public:
  void solve() {
    int t;
    cin >> t;
    while (t--) {
      vector<int> f;
      int n, d;
      cin >> n >> d;
      int N = n, i = 0, sum = 0;
      while (N) {
        if (N <= (1 << i)) {
          f.push_back(N);
          N = 0;
        } else {
          f.push_back(1 << i);
          N -= f[i];
        }
        sum += i * f[i];
        ++i;
      }
      if (sum > d) {
        cout << "NO\n";
        continue;
      }
      d -= sum;
      for (int i = f.size() - 1; i >= 0 && d > 0; --i) {
        while (f[i] != 1 && d) {
          if (d >= f.size() - i) {
            d -= f.size() - i;
            f.push_back(1);
            f[i]--;
          } else {
            f[i + d]++;
            d = 0;
            f[i]--;
          }
        }
      }
      if (d != 0) {
        cout << "NO\n";
        continue;
      }
      cout << "YES\n";
      ind = 1;
      f.push_back(0);
      dfs(1, 0, 0, f);
      for (int i = 2; i <= n; ++i) cout << fa[i] << " ";
      cout << "\n";
    }
  }
  int fa[MAXN], ind = 1;
  void dfs(int u, int ffa, int d, vector<int> &f) {
    fa[u] = ffa;
    if (f[d + 1]) {
      f[d + 1]--;
      dfs(++ind, u, d + 1, f);
    }
    if (f[d + 1]) {
      f[d + 1]--;
      dfs(++ind, u, d + 1, f);
    }
  }
} NSPACE;
int main() {
  ;
  ios_base::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  NSPACE.solve();
}