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;
template <typename T>
void ckMin(T& a, T b) {
  a = min(a, b);
}
template <typename T>
void ckMax(T& a, T b) {
  a = max(a, b);
}
int amt[5000];
int n, d;
bool solve() {
  cin >> n >> d;
  int currSize = n * (n - 1) / 2;
  memset(amt, 0, sizeof(amt));
  for (auto(i) = (0); (i) < (n); ++(i)) amt[i] = 1;
  if (currSize == d) return true;
  for (auto(currD) = (n - 1); (currD) >= (0); --(currD)) {
    while (amt[currD]) {
      amt[currD]--;
      int lowest = currD;
      for (auto(i) = (currD - 1); (i) >= (1); --(i)) {
        if (amt[i] < 2 * amt[i - 1]) {
          lowest = i;
          if (currSize - (currD - i) == d) {
            amt[i]++;
            return true;
          }
        }
      }
      if (lowest == currD) return false;
      currSize -= currD - lowest;
      amt[lowest]++;
    }
  }
  return false;
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  int tc;
  cin >> tc;
  while (tc--) {
    if (solve()) {
      cout << "YES\n";
      int cntTotal = 1;
      for (auto(i) = (1); (i) < (n); ++(i)) {
        for (auto(j) = (0); (j) < (amt[i]); ++(j))
          cout << 1 + cntTotal - amt[i - 1] + j / 2 << ' ';
        cntTotal += amt[i];
      }
      cout << '\n';
    } else {
      cout << "NO\n";
    }
  }
}