#include <bits/stdc++.h>
using namespace std;
string to_string(string s) { return s; }
string to_string(char c) {
  string s = string(1, c);
  return s;
}
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A>
string to_string(vector<A> v) {
  string s = "(";
  int first = 1;
  for (A a : v) {
    if (!first) {
      s += ", ";
    }
    first = 0;
    s += to_string(a);
  }
  s += ")";
  return s;
}
template <typename A>
string to_string(set<A> v) {
  string s = "(";
  int first = 1;
  for (A a : v) {
    if (!first) {
      s += ", ";
    }
    first = 0;
    s += to_string(a);
  }
  s += ")";
  return s;
}
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
const long long inf = (long long)1e18 + 5;
struct leaf {
  long long v, d;
};
bool operator<(const leaf& a, const leaf& b) { return a.d > b.d; };
void solve() {
  long long n, d;
  cin >> n >> d;
  long long l = 0, r = (n * (n - 1)) / 2;
  long long len = -1;
  for (long long i = 1; i < n + 1; i++) {
    if ((i & (i - 1)) == 0) len++;
    l += len;
  }
  if (d < l || d > r) {
    cout << "NO" << endl;
    return;
  }
  priority_queue<leaf> leaves;
  vector<long long> par(n);
  vector<long long> dep(n);
  vector<long long> cnt(n);
  vector<long long> bad(n);
  vector<set<long long>> nodes(n);
  for (long long i = 0; i < n; i++) {
    par[i] = i - 1;
    dep[i] = i;
    cnt[i] = 1;
    nodes[i].insert(i);
  }
  cnt[n - 1] = 0;
  leaves.push({n - 1, n - 1});
  long long cur = (n * (n - 1)) / 2;
  while (cur > d && leaves.size()) {
    leaf at = leaves.top();
    leaves.pop();
    if (at.d - 2 < 0 || nodes[at.d - 2].size() == 0) {
      bad[at.v] = 1;
      continue;
    }
    long long oldp = par[at.v];
    long long p = *nodes[at.d - 2].begin();
    cnt[p]++;
    cnt[oldp]--;
    dep[at.v]--;
    par[at.v] = p;
    if (cnt[p] == 2) nodes[at.d - 2].erase(p);
    if (cnt[oldp] == 1) nodes[at.d - 1].insert(oldp);
    nodes[at.d].erase(at.v);
    nodes[at.d - 1].insert(at.v);
    if (at.d - 1 >= 2) {
      if (!bad[at.v]) leaves.push({at.v, at.d - 1});
      if (!bad[oldp] && cnt[oldp] == 0) leaves.push({oldp, at.d - 1});
    }
    cur--;
  }
  cout << "YES" << endl;
  for (long long i = 1; i < n; i++) {
    cout << par[i] + 1 << " ";
  }
  cout << endl;
}
signed main() {
  long long t = 1;
  cin >> t;
  for (long long i = 0; i < t; i++) solve();
  return 0;
}

### Prompt

Create a solution in cpp for the following 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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
string to_string(string s) { return s; }
string to_string(char c) {
  string s = string(1, c);
  return s;
}
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A>
string to_string(vector<A> v) {
  string s = "(";
  int first = 1;
  for (A a : v) {
    if (!first) {
      s += ", ";
    }
    first = 0;
    s += to_string(a);
  }
  s += ")";
  return s;
}
template <typename A>
string to_string(set<A> v) {
  string s = "(";
  int first = 1;
  for (A a : v) {
    if (!first) {
      s += ", ";
    }
    first = 0;
    s += to_string(a);
  }
  s += ")";
  return s;
}
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
const long long inf = (long long)1e18 + 5;
struct leaf {
  long long v, d;
};
bool operator<(const leaf& a, const leaf& b) { return a.d > b.d; };
void solve() {
  long long n, d;
  cin >> n >> d;
  long long l = 0, r = (n * (n - 1)) / 2;
  long long len = -1;
  for (long long i = 1; i < n + 1; i++) {
    if ((i & (i - 1)) == 0) len++;
    l += len;
  }
  if (d < l || d > r) {
    cout << "NO" << endl;
    return;
  }
  priority_queue<leaf> leaves;
  vector<long long> par(n);
  vector<long long> dep(n);
  vector<long long> cnt(n);
  vector<long long> bad(n);
  vector<set<long long>> nodes(n);
  for (long long i = 0; i < n; i++) {
    par[i] = i - 1;
    dep[i] = i;
    cnt[i] = 1;
    nodes[i].insert(i);
  }
  cnt[n - 1] = 0;
  leaves.push({n - 1, n - 1});
  long long cur = (n * (n - 1)) / 2;
  while (cur > d && leaves.size()) {
    leaf at = leaves.top();
    leaves.pop();
    if (at.d - 2 < 0 || nodes[at.d - 2].size() == 0) {
      bad[at.v] = 1;
      continue;
    }
    long long oldp = par[at.v];
    long long p = *nodes[at.d - 2].begin();
    cnt[p]++;
    cnt[oldp]--;
    dep[at.v]--;
    par[at.v] = p;
    if (cnt[p] == 2) nodes[at.d - 2].erase(p);
    if (cnt[oldp] == 1) nodes[at.d - 1].insert(oldp);
    nodes[at.d].erase(at.v);
    nodes[at.d - 1].insert(at.v);
    if (at.d - 1 >= 2) {
      if (!bad[at.v]) leaves.push({at.v, at.d - 1});
      if (!bad[oldp] && cnt[oldp] == 0) leaves.push({oldp, at.d - 1});
    }
    cur--;
  }
  cout << "YES" << endl;
  for (long long i = 1; i < n; i++) {
    cout << par[i] + 1 << " ";
  }
  cout << endl;
}
signed main() {
  long long t = 1;
  cin >> t;
  for (long long i = 0; i < t; i++) solve();
  return 0;
}

```