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("O2")
using namespace std;
const int MAX = 5e3 + 5;
const long long MAX2 = 11;
const long long MOD = 131071;
const long long MOD2 = 1000000006;
const long long INF = 2e18;
const int dr[] = {1, 0, -1, 0, 1, 1, -1, -1, 0};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1, 0};
const double pi = acos(-1);
const double EPS = 1e-9;
const int block = 316;
int tc, n, k, nw, par[MAX], num[MAX], chl[MAX], fn, dec;
queue<int> z[MAX];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> tc;
  while (tc--) {
    cin >> n >> k;
    nw = n * (n - 1) >> 1;
    if (nw < k) {
      cout << "NO\n";
      continue;
    }
    for (long long i = 1; i <= n; ++i) {
      while (!z[i - 1].empty()) z[i - 1].pop();
      z[i - 1].push(i);
      par[i] = num[i] = i - 1, chl[i - 1] = 1;
    }
    chl[n] = 0;
    for (int i = n; i >= 2; --i) {
      if (chl[i] || chl[par[i]] == 2 || nw == k) break;
      chl[par[i]] = 0;
      ++nw;
      for (int j = i - 2; j >= 0; --j) {
        if (z[j].empty()) break;
        --nw;
        fn = z[j].front();
        if (nw == k) break;
      }
      par[i] = fn;
      chl[fn]++;
      if (chl[fn] == 2) z[num[fn]].pop();
      num[i] = num[fn] + 1;
      z[num[i]].push(i);
    }
    if (nw != k)
      cout << "NO\n";
    else {
      cout << "YES\n";
      for (long long i = 2; i <= n; ++i) cout << par[i] << ' ';
      cout << '\n';
    }
  }
  return 0;
}