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 minans[5005];
int fa[5005];
int dep[5005];
multiset<int> s[5005];
int main() {
  int t, n, d, tmp, dd, p;
  minans[1] = 0;
  tmp = 0;
  for (int i = 2; i <= 5000; i++) {
    if ((i & (-i)) == i) tmp++;
    minans[i] = minans[i - 1] + tmp;
  }
  scanf("%d", &t);
  while (t--) {
    scanf("%d%d", &n, &d);
    if (d > n * (n - 1) / 2) {
      printf("NO\n");
      continue;
    }
    if (d < minans[n]) {
      printf("NO\n");
      continue;
    }
    printf("YES\n");
    for (int i = 2; i <= n; i++) fa[i] = i - 1;
    dd = n * (n - 1) / 2 - d;
    p = 1;
    for (int i = 0; i < n; i++) s[i].insert(i + 1);
    s[n - 1].insert(n);
    for (int i = 1; i <= n; i++) dep[i] = i - 1;
    for (int i = n; i >= 2 && dd > 0; i--)
      if (dd >= (i - 1) - p) {
        s[dep[i]].erase(i);
        s[dep[i] - 1].insert(fa[i]);
        fa[i] = (*s[p - 1].begin());
        s[p - 1].erase(s[p - 1].begin());
        s[p].insert(i);
        s[p].insert(i);
        dep[i] = p;
        dd -= (i - 1) - p;
        while (s[p - 1].empty()) p++;
      } else {
        fa[i] = (*s[(i - 1) - dd - 1].begin());
        break;
      }
    for (int i = 2; i <= n; i++)
      if (i != n)
        printf("%d ", fa[i]);
      else
        printf("%d\n", fa[i]);
    for (int i = 0; i < n; i++) s[i].clear();
  }
  return 0;
}