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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout.setf(ios::fixed);
  cout.precision(15);
  cerr.setf(ios::fixed);
  cerr.precision(10);
  int T;
  cin >> T;
  while (T--) {
    int n, d;
    cin >> n >> d;
    cerr << "n = " << n << ", d = " << d << endl;
    if (d > n * (n - 1) / 2) {
      cout << "NO" << endl;
      continue;
    }
    vector<int> cntChildren(n);
    vector<int> h(n);
    vector<int> p(n);
    for (int i = 1; i < n; i++) {
      p[i] = (i - 1) / 2;
      h[i] = h[p[i]] + 1;
      cntChildren[p[i]]++;
    }
    for (int x : cntChildren) assert(x <= 2);
    int sumH = accumulate((h).begin(), (h).end(), 0);
    if (d < sumH) {
      cout << "NO" << endl;
      continue;
    }
    int remH = d - sumH;
    while (remH > 0) {
      int v1 = -1;
      int v2 = -1;
      for (int i = 0; i < n; i++) {
        if (cntChildren[i] == 0) {
          if (v1 == -1) {
            v1 = i;
          } else if (v2 == -1) {
            v2 = i;
            if (h[v2] > h[v1]) swap(v1, v2);
          } else {
            if (h[i] >= h[v1]) {
              v2 = v1;
              v1 = i;
            } else if (h[i] >= h[v2]) {
              v2 = i;
            }
          }
        }
      }
      assert(v2 != -1);
      assert(cntChildren[v1] == 0);
      assert(cntChildren[v2] == 0);
      auto attach = [&](const int u1, const int u2) {
        assert(h[u1] >= h[u2]);
        cntChildren[p[u2]]--;
        assert(cntChildren[p[u2]] >= 0);
        p[u2] = u1;
        remH += h[u2];
        h[u2] = h[u1] + 1;
        remH -= h[u2];
        cntChildren[u1]++;
      };
      if (remH >= h[v1] + 1 - h[v2]) {
        attach(v1, v2);
      } else {
        int bestV1 = -1;
        while (v1 != 0) {
          v1 = p[v1];
          if (cntChildren[v1] < 2 && remH >= h[v1] + 1 - h[v2] &&
              h[v1] >= h[v2]) {
            bestV1 = v1;
          }
        }
        assert(bestV1 != -1);
        attach(bestV1, v2);
      }
    }
    assert(remH == 0);
    assert(accumulate((h).begin(), (h).end(), 0) == d);
    cout << "YES" << endl;
    for (int i = 1; i < n; i++) {
      cout << (p[i] + 1) << ' ';
    }
    cout << endl;
  }
}