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;
const int N = 5e3 + 5;
int n, d;
int ans[N];
int ch[N];
int fa[N];
int c, now, mx;
int cnt, pos;
int get_dep() {
  int ans = now;
  c++;
  if (c == mx) now++, mx *= 2, c = 1;
  return ans;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    memset(ch, 0, sizeof(ch));
    cin >> n >> d;
    cnt = 0;
    for (int i = 1; i <= n - 1; i++) {
      ans[i] = i;
      cnt += i;
    }
    pos = n - 1;
    now = 1, c = 1, mx = 2;
    if (cnt < d) {
      puts("NO");
      continue;
    }
    while (cnt > d) {
      int dep = get_dep();
      if (dep >= pos) break;
      int tmp = cnt - d;
      if (tmp >= (pos - dep)) {
        cnt -= pos - dep;
        ans[pos] = dep;
      } else {
        ans[pos] -= tmp;
        cnt = d;
      }
      pos--;
    }
    if (cnt > d)
      puts("NO");
    else {
      puts("YES");
      ans[0] = 0;
      for (int i = 1; i <= n - 1; i++) {
        for (int j = 0; j <= n - 1; j++) {
          if (ans[j] == ans[i] - 1 && ch[j] < 2) {
            ch[j]++;
            fa[i] = j;
            break;
          }
        }
      }
      for (int i = 1; i <= n - 1; i++) {
        printf("%d%c", fa[i] + 1, i == n - 1 ? '\n' : ' ');
      }
    }
  }
  return 0;
}