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() {
  int t;
  cin >> t;
  for (int tc = 0; tc < t; tc++) {
    int n, d;
    cin >> n >> d;
    int mnd = 0, mxd = (n * (n - 1)) / 2;
    int p = n - 1, res = 2, k = 1;
    vector<int> taken(5005, 0);
    taken[0] = 1;
    while (p != 0) {
      taken[k] = min(p, res);
      p = p - taken[k];
      mnd = mnd + k * taken[k];
      res = res * 2;
      k++;
    }
    if (d < mnd) {
      cout << "NO\n";
      continue;
    } else if (d > mxd) {
      cout << "NO\n";
      continue;
    } else {
      cout << "YES\n";
    }
    k--;
    int r = mnd;
    while (r != d) {
      for (int i = 1; i <= k; i++) {
        if (2 * (taken[i] - 1) >= taken[i + 1] + 1 && taken[i] > 1) {
          taken[i]--;
          taken[i + 1]++;
          k = max(k, i + 1);
          r++;
          if (r == d) {
            break;
          }
        }
      }
    }
    queue<int> taken3;
    for (int i = 1; i <= 5000; i++) {
      if (taken[i] == 0) {
        break;
      }
      taken3.push(i);
    }
    vector<int> taken2 = taken;
    vector<int> depth(n + 3);
    depth[1] = 0;
    for (int i = 2; i <= n; i++) {
      int u = taken3.front();
      depth[i] = u;
      taken2[u]--;
      if (taken2[u] == 0) {
        taken3.pop();
      }
    }
    queue<int> dep[5005];
    vector<int> cnt(n + 3, 2);
    for (int i = 1; i <= n; i++) {
      dep[depth[i]].push(i);
    }
    for (int i = 2; i <= n; i++) {
      int a = depth[i];
      int x = dep[a - 1].front();
      cout << x << ' ';
      cnt[x]--;
      if (cnt[x] == 0) {
        dep[a - 1].pop();
      }
    }
    cout << '\n';
  }
  return 0;
}