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:

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<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int kMaxN = 5001;
int c[kMaxN], a[kMaxN][kMaxN], f[kMaxN];
int t, n, d;
bool Check() {
  d = n * (n - 1) / 2 - d;
  if (d < 0) {
    return 0;
  }
  for (int i = n - 1, j = 1, k; i > j && d; i--) {
    k = min(i - j, d);
    d -= k;
    k = i - k;
    f[i + 1] = a[k - 1][c[k] / 2];
    a[k][c[k]++] = i + 1;
    if (c[j] >= (1 << j)) {
      j++;
    }
  }
  return !d;
}
int main() {
  cin >> t;
  while (t--) {
    cin >> n >> d;
    for (int i = 0; i < n; i++) {
      c[i] = 1;
      a[i][0] = i + 1;
      f[i + 1] = i;
    }
    if (Check()) {
      cout << "YES" << endl;
      for (int i = 2; i <= n; i++) {
        cout << f[i] << " ";
      }
      cout << endl;
    } else {
      cout << "NO" << endl;
    }
  }
  return 0;
}