#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline bool uax(T &x, T y) {
  return (y > x) ? x = y, true : false;
}
template <typename T>
inline bool uin(T &x, T y) {
  return (y < x) ? x = y, true : false;
}
string to_string(char c) { return "'" + string(1, c) + "'"; }
string to_string(string s) { return '"' + s + '"'; }
string to_string(const char *s) { return to_string((string)s); }
template <typename A>
string to_string(A);
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(A v) {
  bool f = false;
  string r = "{";
  for (auto x : v) {
    if (f) r += ", ";
    r += to_string(x);
    f = true;
  }
  return r += "}";
}
template <typename A>
string to_string(vector<vector<A>> v) {
  string r;
  for (auto x : v) r += "\n" + to_string(x);
  return r;
}
int Nerr;
template <typename A>
string to_string(A *p) {
  return to_string(vector<A>(p, p + Nerr));
}
void err(istream_iterator<string>) { cerr << endl; }
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " =: " << to_string(a) << "; ";
  err(++it, args...);
}
template <typename T>
void kek(T ans) {
  cout << ans << endl;
  exit(0);
}
int const MOD = 1e9 + 7;
long long const INF = 1e18 + 42;
const int N = 5001;
int dq[N], up[N];
int good(int n, int d) { return d >= dq[n] and d <= n * (n - 1) / 2; }
void go(int root, int l, int r, int d) {
  if (l > r) return;
  int n = r - l + 1;
  d -= n;
  if (good(n, d)) {
    up[l] = root;
    return go(l, l + 1, r, d);
  }
  for (int i = 1;; ++i) {
    int j = n - i;
    int li = dq[i], lj = dq[j];
    int ri = i * (i - 1) / 2, rj = j * (j - 1) / 2;
    if (d >= li + lj && d <= ri + rj) {
      up[l] = up[r] = root;
      int di = li, dj = d - di;
      if (dj > rj) {
        dj = rj;
        di = d - dj;
      }
      return go(l, l + 1, l + i - 1, di), go(r, l + i, r - 1, dj);
    }
  }
}
void solve_case() {
  int n, d;
  cin >> n >> d;
  if (!good(n, d)) {
    cout << "NO\n";
    return;
  }
  cout << "YES\n";
  go(0, 1, n - 1, d);
  for (int i = 1; i < n; ++i) {
    cout << up[i] + 1 << ' ';
  }
  cout << '\n';
}
int32_t main() {
  cin.tie(nullptr)->sync_with_stdio(false);
  for (int i = 2; i < N; ++i) {
    dq[i] = MOD;
    for (int j = 0; j < i; ++j) {
      uin(dq[i], dq[j] + j + dq[i - 1 - j] + (i - 1 - j));
    }
  }
  Nerr = 15;
  42;
  ;
  int tt;
  cin >> tt;
  while (tt--) {
    solve_case();
  }
}

### Prompt

Your task is to create a CPP solution to 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;
template <typename T>
inline bool uax(T &x, T y) {
  return (y > x) ? x = y, true : false;
}
template <typename T>
inline bool uin(T &x, T y) {
  return (y < x) ? x = y, true : false;
}
string to_string(char c) { return "'" + string(1, c) + "'"; }
string to_string(string s) { return '"' + s + '"'; }
string to_string(const char *s) { return to_string((string)s); }
template <typename A>
string to_string(A);
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(A v) {
  bool f = false;
  string r = "{";
  for (auto x : v) {
    if (f) r += ", ";
    r += to_string(x);
    f = true;
  }
  return r += "}";
}
template <typename A>
string to_string(vector<vector<A>> v) {
  string r;
  for (auto x : v) r += "\n" + to_string(x);
  return r;
}
int Nerr;
template <typename A>
string to_string(A *p) {
  return to_string(vector<A>(p, p + Nerr));
}
void err(istream_iterator<string>) { cerr << endl; }
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " =: " << to_string(a) << "; ";
  err(++it, args...);
}
template <typename T>
void kek(T ans) {
  cout << ans << endl;
  exit(0);
}
int const MOD = 1e9 + 7;
long long const INF = 1e18 + 42;
const int N = 5001;
int dq[N], up[N];
int good(int n, int d) { return d >= dq[n] and d <= n * (n - 1) / 2; }
void go(int root, int l, int r, int d) {
  if (l > r) return;
  int n = r - l + 1;
  d -= n;
  if (good(n, d)) {
    up[l] = root;
    return go(l, l + 1, r, d);
  }
  for (int i = 1;; ++i) {
    int j = n - i;
    int li = dq[i], lj = dq[j];
    int ri = i * (i - 1) / 2, rj = j * (j - 1) / 2;
    if (d >= li + lj && d <= ri + rj) {
      up[l] = up[r] = root;
      int di = li, dj = d - di;
      if (dj > rj) {
        dj = rj;
        di = d - dj;
      }
      return go(l, l + 1, l + i - 1, di), go(r, l + i, r - 1, dj);
    }
  }
}
void solve_case() {
  int n, d;
  cin >> n >> d;
  if (!good(n, d)) {
    cout << "NO\n";
    return;
  }
  cout << "YES\n";
  go(0, 1, n - 1, d);
  for (int i = 1; i < n; ++i) {
    cout << up[i] + 1 << ' ';
  }
  cout << '\n';
}
int32_t main() {
  cin.tie(nullptr)->sync_with_stdio(false);
  for (int i = 2; i < N; ++i) {
    dq[i] = MOD;
    for (int j = 0; j < i; ++j) {
      uin(dq[i], dq[j] + j + dq[i - 1 - j] + (i - 1 - j));
    }
  }
  Nerr = 15;
  42;
  ;
  int tt;
  cin >> tt;
  while (tt--) {
    solve_case();
  }
}

```