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() {
  int t;
  cin >> t;
  while (t--) {
    int n, d, i, j, fl = 0, gl = 0, l, o = 0;
    cin >> n >> d;
    int depth[n + 1];
    int res = n * (n - 1) / 2;
    vector<vector<int>> v(n + 1), dp(n);
    if (d > n * (n - 1) / 2) {
      cout << "NO" << endl;
      o = 1;
    } else {
      for (i = 1; i <= n; i++) depth[i] = i - 1;
      for (i = 1; i < n; i++) v[i].push_back(i + 1);
      for (i = 0; i < n; i++) dp[i].push_back(i + 1);
      while (1) {
        for (i = 1; i <= n; i++) {
          if (v[i].size() == 0) {
            for (j = 0; j < n; j++) {
              for (int h : dp[j]) {
                if (j + 1 >= depth[i]) break;
                if ((res - depth[i] + depth[h] + 1) < d || j > depth[i]) break;
                if (v[h].size() < 2 && h != i) {
                  int s = h;
                  res = res - depth[i] + depth[h] + 1;
                  v[h].push_back(i);
                  for (l = 1; l <= n; l++) {
                    if (l != s) {
                      auto it = find(v[l].begin(), v[l].end(), i);
                      if (it != v[l].end()) {
                        v[l].erase(it);
                        break;
                      }
                    }
                  }
                  auto it = find(dp[depth[i]].begin(), dp[depth[i]].end(), i);
                  dp[depth[i]].erase(it);
                  dp[depth[s] + 1].push_back(i);
                  depth[i] = depth[s] + 1;
                  gl = 1;
                  break;
                }
              }
              if (gl == 1) {
                gl = 0;
                break;
              }
            }
            if (j < n) break;
          }
        }
        if (res == d || i == n + 1) break;
      }
    }
    if (o == 0) {
      if (res == d) {
        cout << "YES" << endl;
        for (i = 2; i <= n; i++) {
          for (j = 1; j <= n; j++) {
            auto it = find(v[j].begin(), v[j].end(), i);
            if (it != v[j].end()) {
              cout << j << " ";
            }
          }
        }
        cout << endl;
      } else
        cout << "NO" << endl;
    }
  }
}