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>
#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, d;
    int64_t cur_d = 0;
    cin >> n >> d;
    vector<int> tree(n);
    vector<int> depth(n, 0);
    vector<vector<int>> children(n);
    queue<int> roots{{0}};
    int cur_leaf = n - 1;
    for (int i = 1; i < n; i++) {
      tree[i] = i - 1;
      children[i - 1].push_back(i);
      cur_d += i;
      depth[i] = i;
    }
    while (cur_d > d) {
      int cur_root = roots.front();
      if (depth[cur_leaf] <= depth[cur_root]) break;
      if (cur_d - (depth[cur_leaf] - depth[cur_root]) >= d) {
        cur_d -= (depth[cur_leaf] - depth[cur_root] - 1);
        depth[cur_leaf] = depth[cur_root] + 1;
        children[cur_root].push_back(cur_leaf);
        tree[cur_leaf] = cur_root;
        children[cur_leaf - 1].pop_back();
        if (children[cur_root].size() >= 2) {
          roots.pop();
          for (int i = children[cur_root].size() - 1; i >= 0; i--) {
            roots.push(children[cur_root][i]);
          }
        }
        cur_leaf--;
        continue;
      } else {
        if (children[cur_leaf - 1].size() >= 2) break;
        depth[cur_leaf]--;
        tree[cur_leaf]--;
        cur_d--;
      }
    }
    if (cur_d != d) {
      cout << "NO\n";
      continue;
    }
    cout << "YES\n";
    for (int i = 1; i < n; i++) {
      cout << tree[i] + 1 << ' ';
    }
    cout << '\n';
  }
  return 0;
}
