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;
long long n, d;
long long N = 5003;
vector<long long> a(4 * N, 1);
long long tree[4 * 5003] = {0};
long long cur = 1;
void solve() {
  a = vector<long long>(4 * N, 1);
  long long tree[4 * 5003] = {0};
  cin >> n >> d;
  long long sum = 0, step = 1, n1 = n;
  for (int i = 0; i < 14; ++i) {
    sum += min(n1, step) * i;
    n1 -= min(n1, step);
    step *= 2;
  }
  if ((n * (n - 1)) / 2 < d or d < sum) {
    cout << "NO\n";
    return;
  }
  long long cur = (n * (n - 1)) / 2;
  for (int i = n - 1; i > -1; i--) {
    for (int j = 1; j < i; ++j) {
      if (a[j - 1] * 2 > a[j] and cur - (i - j) >= d) {
        a[j]++;
        a[i]--;
        cur -= (i - j);
        break;
      }
      if (cur == d) break;
    }
    if (cur == d) break;
  }
  step = 0;
  long long l = 0, r = 0;
  long long curN = 1;
  long long curr;
  long long N = 1;
  for (int i = 0; i < n; ++i) {
    curN = a[i];
    long long curr = l;
    while (curN > 0) {
      if (curN == 1) {
        tree[N] = curr;
        curN = 0;
        N += 1;
      } else {
        tree[N] = curr;
        tree[N + 1] = curr;
        curr += 1;
        curN -= 2;
        N += 2;
      }
    }
    l = r + 1;
    r = l + a[i] - 1;
  }
  cout << "YES\n";
  for (int i = 2; i <= n; ++i) {
    cout << tree[i] << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  for (int i = 0; i < t; ++i) {
    solve();
  }
}