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>
const long long mod = 1000000007, N = 5004;
using namespace std;
long long n, m, i, j, k, l, r, t, ans, tt;
long long p[N], d[N], ava[N];
long long v[N];
int main() {
  cin >> tt;
  while (tt--) {
    cin >> n >> m;
    if (m > n * (n - 1) / 2) {
      cout << "NO\n";
      continue;
    }
    k = n * (n - 1) / 2;
    for (i = 1; i <= n; ++i) {
      p[i] = i - 1;
      d[i] = i;
      ava[i] = 1;
    }
    ava[n] = 2;
    while (k > m) {
      t = 0;
      d[0] = 0;
      for (i = 1; i <= n; ++i) {
        if (d[i] > d[t]) t = i;
      }
      l = 0;
      d[0] = mod;
      for (i = 1; i <= n; ++i) {
        if (d[i] < d[l] && ava[i] > 0) l = i;
      }
      r = d[l] + 1;
      if (d[t] - r >= k - m) {
        for (i = 1; i <= n; ++i) {
          if (d[t] - d[i] - 1 == k - m && ava[i] > 0) {
            p[t] = i;
            d[t] = d[i] + 1;
            break;
          }
        }
        break;
      }
      if (r >= d[t]) break;
      k -= (d[t] - r);
      d[t] = r;
      ava[p[t]]++;
      p[t] = l;
      ava[l]--;
      ava[t] = 2;
    }
    t = 0;
    for (i = 1; i <= n; ++i) t += d[i] - 1;
    if (t != m) {
      cout << "NO\n";
      continue;
    }
    cout << "YES\n";
    for (i = 2; i <= n; ++i) cout << p[i] << " ";
    cout << endl;
  }
  return 0;
}