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;
template <typename T>
using pq_asc = priority_queue<T, vector<T>, greater<T>>;
template <typename T>
ostream &JOUT(T s, T e, string sep = " ", ostream &os = cout) {
  if (s != e) {
    os << *s;
    ++s;
  }
  while (s != e) {
    os << sep << *s;
    ++s;
  }
  return os;
}
ostream &YES(bool cond, string yes = "Yes", string no = "No",
             ostream &os = cout) {
  if (cond) {
    os << yes << endl;
  } else {
    os << no << endl;
  }
  return os;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) {
  os << '(' << p.first << ", " << p.second << ')';
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << '[';
  JOUT((v).begin(), (v).end(), ", ", os) << ']';
  return os;
}
const long long INF = 1e18;
const long long MOD = 1e9 + 7;
void solve() {
  long long n;
  cin >> n;
  ;
  long long d;
  cin >> d;
  ;
  long long s = n * (n - 1) / 2;
  if (s < d) {
    cout << ("NO") << endl;
    ;
    return;
  }
  vector<long long> c(n, 1);
  long long k = 1;
  for (long long i = n - 1; i >= 1; --i) {
    --c[i];
    long long j = max(k, i - (s - d));
    ++c[j];
    s -= i - j;
    if (s == d) {
      break;
    }
    if (c[k] == (1LL << k)) {
      ++k;
    }
  }
  if (s > d) {
    cout << ("NO") << endl;
    ;
    return;
  }
  vector<long long> acc(n + 1);
  for (long long i = 0; i < (n); ++i) {
    acc[i + 1] = acc[i] + c[i];
  }
  vector<long long> ans(n - 1);
  long long j = 1;
  long long p = 1;
  long long q = 0;
  for (long long i = 0; i < (n - 1); ++i) {
    ans[i] = p;
    --c[j];
    if (c[j] == 0) {
      p = acc[j] + 1;
      q = 0;
      ++j;
    } else {
      ++q;
      if (q == 2) {
        ++p;
        q = 0;
      }
    }
  }
  cout << ("YES") << endl;
  ;
  JOUT((ans).begin(), (ans).end()) << endl;
}
int main() {
  long long t;
  cin >> t;
  ;
  for (long long i = 0; i < (t); ++i) {
    solve();
  }
}