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 long long mod = 1e18;
const long long inf = 1e18;
const double eps = 1e-8;
const double pi = acos(-1);
int p[5020];
int dp[5030];
vector<int> v[5030];
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, d;
    cin >> n >> d;
    if (d > (n * (n - 1) / 2)) {
      cout << "NO\n";
      continue;
    }
    int dep = 0;
    for (int i = 2; i < n + 1; i++) {
      p[i] = i - 1;
      dp[i] = i - 1;
      dep += i - 1;
    }
    int k = n, tmp = 0;
    v[tmp].push_back(1);
    int flog = 0;
    while (dep > d) {
      if (k == 1) {
        flog = 1;
        break;
      }
      if (v[tmp].size() == (1 << tmp)) {
        tmp++;
        v[tmp].push_back(tmp + 1);
      }
      if (dp[k] - tmp > dep - d) {
        tmp = dp[k] - (dep - d);
        if (!v[tmp].size()) v[tmp].push_back(tmp + 1);
        if (!v[tmp - 1].size()) v[tmp - 1].push_back(tmp);
      }
      int sz = v[tmp].size();
      v[tmp].push_back(k);
      dep -= dp[k];
      dep += tmp;
      dp[k] = tmp;
      p[k] = v[tmp - 1][sz / 2];
      k--;
    }
    if (flog) {
      cout << "NO\n";
      for (int i = 0; i < tmp + 3; i++) v[i].clear();
      continue;
    }
    cout << "YES\n";
    for (int i = 2; i < n + 1; i++) cout << p[i] << ' ';
    cout << '\n';
    for (int i = 0; i < tmp + 3; i++) v[i].clear();
  }
  return 0;
}