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 mod = 1000000007;
const int mj = 2e5 + 5;
template <typename T>
T add(T& a, T b) {
  a += b;
  while (a >= mod) a -= mod;
  while (a < 0) a += mod;
}
template <typename T>
T mul(T& a, T b) {
  a = a * b % mod;
}
template <typename T>
T up_self(T& a, T b) {
  a = max(a, b);
}
template <typename T>
T min_self(T& a, T b) {
  a = min(a, b);
}
string find_largest_pref_palindrome(const string& a) {
  string second = a;
  reverse(second.begin(), second.end());
  second = a + "%" + second;
  int c = 0;
  vector<int> pref(2 * ((int)a.size()) + 2);
  for (int i = 1; i < second.size(); i++) {
    while (c != 0 && second[i] != second[c]) c = pref[c - 1];
    if (second[i] == second[c]) c++;
    pref[i] = c;
  }
  return second.substr(0, c);
}
long long int binexpomodulo(long long int x, long long int y) {
  long long int res = 1;
  x %= mod;
  if (!x) return 0;
  while (y) {
    if (y & 1) {
      mul(res, x);
    }
    mul(x, x);
    y >>= 1;
  }
  return res;
}
long long int nCrInOr(long long int n, long long int r) {
  long long int res = 1;
  if (r > n - r) r = n - r;
  long long int rin = 1;
  for (long long int i = 1; i <= r; i++) rin = (rin * i) % mod;
  rin = binexpomodulo(rin, mod - 2);
  for (long long int i = 1; i <= r; i++) res = (res * (n - i + 1)) % mod;
  res = (res * rin) % mod;
  return res;
}
int msb(long long int n) {
  int ans;
  for (int i = 0; i < 64; i++)
    if (n & (1LL << i)) ans = i + 1;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t;
  cin >> t;
  while (t--) {
    int n, d;
    cin >> n >> d;
    int ld = 0, ud = 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 <= ud)
      cout << "YES"
           << "\n";
    else {
      cout << "NO"
           << "\n";
      continue;
    }
    vector<int> par(n), dep(n), bad(n), cnt(n, 1);
    iota(par.begin(), par.end(), -1);
    iota(dep.begin(), dep.end(), 0);
    cnt[n - 1] = 0;
    int curr = ud;
    while (curr > d) {
      int v = -1;
      for (int i = 0; i < n; i++) {
        if (!bad[i] && cnt[i] == 0 && (v == -1 || dep[v] > dep[i])) v = i;
      }
      int p = -1;
      for (int i = 0; i < n; i++) {
        if (cnt[i] < 2 && dep[i] < dep[v] - 1 && (p == -1 || dep[p] < dep[i]))
          p = i;
      }
      if (p == -1) {
        bad[v] = 1;
        continue;
      }
      --cnt[par[v]];
      par[v] = p;
      ++cnt[p];
      --curr;
      --dep[v];
    }
    for (int i = 1; i < n; i++) cout << par[i] + 1 << " ";
    cout << "\n";
  }
  return 0;
}