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>
#pragma GCC optimize("O3")
using namespace std;
const int mod = 1000000007;
const double dancila = 3.14159265359;
const double eps = 1e-9;
int t;
int frq[5002], tt[5005], x = 1;
void dfs(int nod, int nivel) {
  if (frq[nivel + 1] > 0) {
    ++x;
    --frq[nivel + 1];
    tt[x] = nod;
    dfs(x, nivel + 1);
  }
  if (frq[nivel + 1] > 0) {
    ++x;
    --frq[nivel + 1];
    tt[x] = nod;
    dfs(x, nivel + 1);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> t;
  for (; t; --t) {
    int n, ad;
    cin >> n >> ad;
    int sum = 0;
    for (int i = 0; i < n; ++i) frq[i] = 1, sum += i;
    for (int i = n - 1; sum > ad && i >= 1; --i)
      while (frq[i] && frq[i] >= (frq[i + 1] / 2) + (frq[i + 1] % 2) &&
             sum > ad) {
        bool ok = 0;
        for (int j = 1; j < i; ++j) {
          if (sum - i + j >= ad && frq[j] + 1 <= 2 * frq[j - 1]) {
            frq[i]--;
            frq[j]++;
            sum = sum - i + j;
            ok = 1;
            break;
          }
        }
        if (!ok) break;
      }
    if (sum != ad) {
      cout << "NO\n";
    } else {
      cout << "YES\n";
      x = 1;
      dfs(1, 0);
      for (int i = 2; i <= n; ++i) cout << tt[i] << " ";
      cout << '\n';
    }
  }
  return 0;
}