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 N = 5003;
int L[N], U[N];
int lb(int n) {
  if (n <= 1) return 0;
  if (L[n]) return L[n];
  int x = n / 2;
  int y = n - x - 1;
  return L[n] = x + y + lb(x) + lb(y);
}
int ub(int n) { return n * (n - 1) / 2; }
void prep() {
  for (int i = 2; i < N; i++) {
    L[i] = lb(i);
    U[i] = ub(i);
  }
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
void solve() {
  int n, d;
  cin >> n >> d;
  if (L[n] > d || d > U[n]) {
    cout << "NO\n";
    return;
  }
  cout << "YES\n";
  vector<int> pa(n);
  int cnt = 1;
  function<void(int, int, int)> divi = [&](int u, int n, int d) {
    if (n <= 1) return;
    int x, dx, y, dy;
    while (true) {
      x = rng() % n;
      dx = L[x] + rng() % (U[x] - L[x] + 1);
      y = n - 1 - x;
      dy = d - (n - 1) - dx;
      if (L[y] <= dy && dy <= U[y]) break;
    }
    if (x) {
      pa[cnt++] = u;
      divi(cnt - 1, x, dx);
    }
    if (y) {
      pa[cnt++] = u;
      divi(cnt - 1, y, dy);
    }
  };
  divi(0, n, d);
  for (int i = 1; i < n; i++) {
    cout << pa[i] + 1 << ' ';
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  prep();
  int T;
  cin >> T;
  for (int t = 0; t < T; t++) {
    solve();
  }
  return 0;
}