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>
#pragma GCC optimize("O3")
using namespace std;
const int N = 5e3 + 10;
const int inf = 1e18;
const int mod = 1e9 + 7;
const int eps = 1e-9;
const char spc = 29;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
using namespace std;
int q;
int n, s;
int d[N];
int p[N];
int mxcnt(int curplaced, int curd, int cnt) {
  int res = 0;
  while (cnt) {
    cnt -= curplaced;
    res += curd * curplaced;
    curplaced = min(curplaced * 2, cnt);
    ++curd;
  }
  return res;
}
int binsearch(int need, int curd, int cnt, int prev) {
  int l = 1;
  int r = min(cnt, prev * 2);
  while (l < r) {
    int mid = (l + r) / 2;
    if (mxcnt(mid, curd, cnt) < need)
      r = mid;
    else
      l = mid + 1;
  }
  if (l > 1 && mxcnt(l - 1, curd, cnt) == need) return l - 1;
  return l;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> q;
  while (q--) {
    cin >> n >> s;
    fill(d, d + n, 0);
    int prev = 1;
    d[0] = 1;
    int cnt = n - 1;
    for (int i = 1; i < n; ++i) {
      if (!cnt) break;
      d[i] = binsearch(s, i, cnt, d[i - 1]);
      cnt -= d[i];
      s -= d[i] * i;
    }
    if (s != 0) {
      cout << "NO"
           << "\n";
      continue;
    }
    vector<int> v;
    v.push_back(0);
    int curd = 1;
    int curid = 1;
    while (d[curd]) {
      vector<int> nv;
      for (int i = 0; i < v.size(); ++i) {
        if (!d[curd]) break;
        nv.push_back(curid);
        p[curid++] = v[i];
        --d[curd];
        if (!d[curd]) break;
        nv.push_back(curid);
        p[curid++] = v[i];
        --d[curd];
      }
      ++curd;
      v = nv;
    }
    cout << "YES"
         << "\n";
    for (int i = 1; i < n; ++i) cout << (p[i] + 1) << " ";
    cout << "\n";
  }
}