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 int inf = 1e9 + 7;
const long long INF = 1e18;
template <typename T1, typename T2>
inline bool chmin(T1& a, T2 b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <typename T1, typename T2>
inline bool chmax(T1& a, T2 b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <typename T>
T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <typename T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
template <typename T>
T pow(T a, int b) {
  return b ? pow(a * a, b / 2) * (b % 2 ? a : 1) : 1;
}
const int mod = 1000000007;
long long modpow(long long a, int b) {
  return b ? modpow(a * a % mod, b / 2) * (b % 2 ? a : 1) % mod : 1;
}
template <class T>
ostream& operator<<(ostream& os, const vector<T>& vec) {
  for (int i = 0; i < ((int)vec.size()); ++i) {
    if (i) os << " ";
    os << vec[i];
  }
  return os;
}
template <class T, class U>
ostream& operator<<(ostream& os, const pair<T, U>& p) {
  os << p.first << " " << p.second;
  return os;
}
template <class T>
inline void add(T& a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
void solve();
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int T;
  cin >> T;
  while (T--) {
    solve();
  }
}
void solve() {
  int n, d;
  cin >> n >> d;
  int ld = 0, rd = n * (n - 1) / 2;
  for (int i = 1, cd = 0; i <= n; ++i) {
    if (!(i & (i - 1))) ++cd;
    ld += cd - 1;
  }
  if (!(ld <= d && d <= rd)) {
    cout << "NO" << endl;
    return;
  }
  vector<int> par(n, -1);
  for (int i = 1; i < (n); ++i) par[i] = i - 1;
  vector<int> c(n, 1);
  c[n - 1] = 0;
  vector<int> bad(n);
  vector<int> dep(n);
  for (int i = 0; i < (n); ++i) dep[i] = i;
  int cur = n * (n - 1) / 2;
  while (cur > d) {
    int v = -1;
    for (int i = 0; i < (n); ++i) {
      if (!bad[i] && c[i] == 0 && (v == -1 || dep[v] > dep[i])) {
        v = i;
      }
    }
    int p = -1;
    for (int i = 0; i < (n); ++i) {
      if (c[i] < 2 && dep[i] == dep[v] - 2) {
        p = i;
      }
    }
    if (p == -1) {
      bad[v] = 1;
      continue;
    }
    --c[par[v]];
    --dep[v];
    ++c[p];
    par[v] = p;
    --cur;
  }
  cout << "YES" << endl;
  for (int i = 1; i < (n); ++i) cout << par[i] + 1 << " ";
  cout << endl;
}