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 long long N = 5005;
long long n, d, hi[N], par[N];
vector<long long> a[N];
queue<long long> leaf;
bool res;
void in() {
  cin >> n >> d;
  for (long long i = 1; i <= n; i++) a[i].clear();
  for (long long i = 1; i <= n; i++) hi[i] = 0;
  while (!leaf.empty()) leaf.pop();
  for (long long i = 1; i <= n; i++) par[i] = 0;
  for (long long i = 1; i <= n; i++) hi[i] = i - 1;
  for (long long i = 1; i <= n - 1; i++) a[i].push_back(i + 1);
  res = false;
  leaf.push(n);
}
void process() {
  long long val = n * (n - 1) / 2;
  while (!leaf.empty()) {
    long long t = leaf.front();
    leaf.pop();
    long long Par;
    if (val == d) {
      res = true;
      break;
    }
    for (long long i = 1; i <= n; i++)
      for (auto j : a[i])
        if (j == t) Par = i;
    long long ver, Min = 1e9;
    for (long long i = 1; i <= n; i++) {
      if (a[i].size() < 2 && val - (hi[t] - hi[i] - 1) >= d) {
        if (Min > hi[i]) {
          Min = hi[i], ver = i;
        }
      }
    }
    if (Min == 1e9) continue;
    a[ver].push_back(t);
    for (long long i = 0; i <= a[Par].size() - 1; i++) {
      if (a[Par][i] == t) {
        a[Par].erase(a[Par].begin() + i);
        break;
      }
    }
    if (a[Par].size() == 0) leaf.push(Par);
    val -= (hi[t] - hi[ver] - 1);
    hi[t] = hi[ver] + 1;
    if (val == d) {
      res = true;
      break;
    }
  }
}
int main() {
  if (fopen("trash.inp", "r"))
    freopen("trash.inp", "r", stdin), freopen("trash.out", "w", stdout);
  ios ::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long T;
  cin >> T;
  while (T--) {
    in();
    if (n * (n - 1) / 2 < d) {
      cout << "NO" << '\n';
      continue;
    }
    process();
    if (!res) {
      cout << "NO" << '\n';
      continue;
    } else {
      cout << "YES" << '\n';
      for (long long i = 1; i <= n; i++)
        for (auto j : a[i]) {
          par[j] = i;
        }
      for (long long i = 2; i <= n; i++) cout << par[i] << ' ';
      cout << '\n';
    }
  }
}