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>
int MinD(int n) {
  std::vector<int> d(n);
  d[0] = 0;
  for (int i = 1; i < n; ++i) {
    d[i] = d[(i - 1) / 2] + 1;
  }
  return std::accumulate(d.begin(), d.end(), 0);
}
int MaxD(int n) { return n * (n - 1) / 2; }
const int N = 668;
const int D = 5000;
int min_d[N + 1], max_d[N + 1];
std::pair<int, int> from[N + 1][D + 1];
int ans[N + 1];
void FillAns(int n, int d, int u, int p) {
  ans[u] = p;
  if (n > 0) {
    int n1 = from[n][d].first;
    int n2 = n - 1 - n1;
    int d1 = from[n][d].second;
    int d2 = d - n1 - n2 - d1;
    FillAns(n1, d1, u + 1, u);
    FillAns(n2, d2, u + n1 + 1, u);
  }
}
int main() {
  for (int i = 1; i <= N; ++i) {
    min_d[i] = MinD(i);
    max_d[i] = MaxD(i);
  }
  for (int i = 0; i <= N; ++i) {
    for (int j = 0; j <= D; ++j) {
      from[i][j] = std::make_pair(-1, -1);
    }
  }
  from[0][0] = std::make_pair(0, 0);
  for (int n = 1; n <= N; ++n) {
    for (int d = min_d[n]; d <= std::min(D, max_d[n]); ++d) {
      for (int n1 = 0; n1 < n; ++n1) {
        int n2 = n - 1 - n1;
        int d_sum = d - n1 - n2;
        if (min_d[n1] + min_d[n2] <= d_sum && d_sum <= max_d[n1] + max_d[n2]) {
          int d1 = std::min(d_sum - min_d[n2], max_d[n1]);
          from[n][d] = std::make_pair(n1, d1);
          break;
        }
      }
    }
  }
  int t;
  std::cin >> t;
  while (t--) {
    int n, d;
    std::cin >> n >> d;
    if (n > N || d < min_d[n] || d > max_d[n]) {
      std::cout << "NO" << std::endl;
      continue;
    }
    FillAns(n, d, 0, -1);
    std::cout << "YES" << std::endl;
    for (int i = 1; i < n; ++i) {
      std::cout << ans[i] + 1 << ' ';
    }
    std::cout << std::endl;
  }
  return 0;
}