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>
const int INF = 0x3f3f3f3f;
using namespace std;
int v[5005];
int main() {
  int t1;
  cin >> t1;
  for (int t = 1; t <= t1; t++) {
    int n, d;
    cin >> n >> d;
    memset(v, 0, sizeof(v));
    int l = 1, nr = n - 1;
    v[0] = 1;
    for (; nr >= v[l - 1] * 2; l++) {
      v[l] = v[l - 1] * 2;
      d -= v[l] * l;
      nr -= v[l];
    }
    if (nr == 0)
      l--;
    else
      d -= nr * l, v[l] = nr;
    if (d < 0) {
      cout << "NO\n";
      continue;
    }
    int l2;
    for (l2 = l; d > 0 && l >= 1; l--) {
      if (v[l] == 1) continue;
      if (l2 + 1 - l > d) {
        v[l]--;
        v[l + d]++;
        d = 0;
        break;
      }
      v[++l2] = 1;
      d -= l2 - l;
      v[l]--;
      l++;
    }
    if (d == 0) {
      cout << "YES\n";
      queue<int> p[5005];
      p[0].push(1);
      p[0].push(1);
      for (int i = 1, node = 2; i <= l2; i++)
        if (v[i] > 0) {
          cout << p[i - 1].front() << ' ';
          p[i - 1].pop();
          p[i].push(node);
          p[i].push(node);
          v[i--]--;
          node++;
        }
      cout << '\n';
    } else {
      cout << "NO\n";
    }
  }
  return 0;
}