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;
const int mod = 1e9 + 7;
long double PI = acosl(-1);
const long long infl = 3e18 + 10;
const int inf = 2e9 + 10;
const int nmax = 5e3 + 10;
const int MAXLG = log2(nmax) + 1;
int lo[nmax], hi[nmax];
int pp[nmax];
int cur;
bool dfs(int u, int nodbaki, int dbaki) {
  if (u == 0) return true;
  if (dbaki > hi[nodbaki] or dbaki < lo[nodbaki]) return false;
  dbaki -= nodbaki - 1;
  if (nodbaki == 1) return true;
  for (int i = 0; i <= (nodbaki - 1) / 2; i++) {
    int bame = i, dane = nodbaki - 1 - bame;
    if (lo[bame] + lo[dane] <= dbaki and hi[bame] + hi[dane] >= dbaki) {
      int v1 = 0, v2 = 0;
      if (bame) v1 = ++cur, pp[v1] = u;
      int z = dbaki - lo[bame] - lo[dane];
      int k1 = min(hi[bame], lo[bame] + z);
      int k2 = dbaki - k1;
      dfs(v1, bame, k1);
      if (dane) v2 = ++cur, pp[v2] = u;
      dfs(v2, dane, k2);
      return true;
    }
  }
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  for (int i = 2; i < nmax; i++) {
    lo[i] = lo[i - 1] + (31 - __builtin_clz(i));
    hi[i] = (i - 1) * i / 2;
  }
  int tc;
  cin >> tc;
  while (tc--) {
    int n, d;
    cin >> n >> d;
    cur = 1;
    bool hobenaki = dfs(1, n, d);
    if (hobenaki) {
      cout << "YES\n";
      for (int i = 2; i <= n; i++) {
        cout << pp[i] << " ";
      }
      cout << "\n";
    } else
      cout << "NO\n";
  }
}