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>
#include <bits/stdc++.h>
using namespace std;
long long N, D, T;
int v[5015];
int v2[5015];
int v3[5015];
void pre() {
  for (int i = 1; (i) <= int(5010); (i)++) {
    v[i + 1] = (i * (i + 1)) / 2;
    v3[i + 1] = i;
  }
  int cont, cont2 = 1, x = 0;
  int valor = 0;
  while (cont2 < 5010) {
    cont = 0;
    valor++;
    int j = 1 << ++x;
    while (cont2++ < 5010 && cont++ < j) {
      v2[cont2] = v2[cont2 - 1] + valor;
    }
    cont2--;
  }
}
void solve() {
  if (D > v[N] || D < v2[N]) {
    cout << "NO"
         << "\n";
    return;
  }
  cout << "YES"
       << "\n";
  multiset<int> ms;
  for (int i = 1; (i) <= int(N); (i)++) {
    ms.insert(v3[i]);
  }
  int cont = v[N];
  while (cont > D) {
    auto it = ms.end();
    it--;
    int aux = *it;
    ms.erase(it);
    int h = 1;
    while (ms.count(h) >= (1 << h)) {
      h++;
    }
    while (cont > D && aux > h) {
      cont--;
      aux--;
    }
    ms.insert(aux);
  }
  int aux;
  vector<int> sol;
  vector<int> altura(N, 0);
  cont = 0;
  for (auto i : ms) {
    cont++;
    if (!i) continue;
    if (!altura[i]) {
      altura[i] = cont;
    }
  }
  for (auto i : ms) {
    if (!i)
      continue;
    else {
      if (i == 1) {
        sol.push_back(1);
        aux = 1;
      } else {
        if (i != aux) cont = 0;
        aux = i;
        int temp = altura[i - 1];
        for (int j = 0; j < cont / 2; j++) {
          temp++;
        }
        sol.push_back(temp);
        cont++;
      }
    }
  }
  for (int i = 0; (i) < int(sol.size() - 1); (i)++) {
    cout << sol[i] << " ";
  }
  cout << sol[(long long)(sol.size()) - 1] << "\n";
}
int main() {
  cin >> T;
  pre();
  while (T--) {
    cin >> N >> D;
    solve();
  }
  return 0;
}
