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 n, d;
long long minLen[5010], maxLen[5010];
int ans[5010];
int prevNum;
bool f(int n, int d, int startNum) {
  ans[startNum] = prevNum;
  if (n == 1) return true;
  for (int p1 = 0; p1 <= n / 2; p1++) {
    int p2 = n - 1 - p1;
    int min1 = minLen[p1];
    int max1 = maxLen[p1];
    int min2 = minLen[p2];
    int max2 = maxLen[p2];
    if (d - n + 1 >= min1 + min2 && d - n + 1 <= max1 + max2) {
      int q1 = min1;
      int q2 = min2;
      int left = (d - n + 1) - (min1 + min2);
      if (left < max1 - min1)
        q1 += left;
      else {
        q1 = max1;
        q2 += (left - (max1 - min1));
      }
      prevNum = startNum;
      f(p1, q1, startNum + 1);
      prevNum = startNum;
      f(p2, q2, startNum + p1 + 1);
      return true;
    }
  }
  return false;
}
int main() {
  minLen[1] = 0;
  int idx = 2;
  for (int i = 1; i < 20; i++) {
    for (int j = 0; j < ((1) << i); j++) {
      if (idx > 5004) break;
      minLen[idx] = minLen[idx - 1] + i;
      idx++;
    }
  }
  for (int i = 0; i < 5010; i++) {
    maxLen[i] = i * (i - 1) / 2;
  }
  int t;
  cin >> t;
  while (t--) {
    cin >> n >> d;
    bool can = f(n, d, 1);
    if (!can) {
      cout << "NO\n";
    } else {
      cout << "YES\n";
      for (int i = 2; i <= n; i++) printf("%d ", ans[i]);
      printf("\n");
    }
  }
}