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:

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<image>
#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int maxn = 5e3 + 10;
const int inf = 0x3f3f3f3f;
int fa[maxn], dep[maxn];
bool check(int x) {
  while (x % 2 == 0) {
    x >>= 1;
  }
  if (x == 1)
    return true;
  else
    return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  dep[1] = 0;
  for (int i = (2); i <= (5000); i++) dep[i] = dep[i >> 1] + 1;
  while (t--) {
    int n, d;
    long long ans = 0;
    cin >> n >> d;
    for (int i = (2); i <= (n); i++) {
      fa[i] = (i >> 1);
      ans += dep[i];
    }
    if (ans > d) {
      cout << "NO" << '\n';
      continue;
    }
    int now = 1;
    while (ans < d && now < n) {
      fa[n - now + 1] = n - now;
      if (check(n - now + 1)) {
        now++;
        continue;
      }
      ans += now;
      now++;
    }
    if (ans < d) {
      cout << "NO" << '\n';
      continue;
    }
    if (ans > d) fa[n] = n - (ans - d) - 1;
    cout << "YES" << '\n';
    for (int i = (2); i <= (n); i++) cout << fa[i] << (i == n ? '\n' : ' ');
  }
  return 0;
}
