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 maxn = 5000 + 5;
int fa[maxn];
set<int> rest;
int main() {
  int w = 2;
  for (int i = 1; i <= 20; i++) {
    rest.insert(w - 1);
    w *= 2;
  }
  int T;
  scanf("%d", &T);
  while (T--) {
    int n, d;
    scanf("%d%d", &n, &d);
    for (int i = 1; i <= n; i++) fa[i] = i - 1;
    int sumdep = (0 + n - 1) * n / 2;
    int nowpos = 1, maxmove = n - 2, twice = 1;
    while (maxmove > 0 && sumdep > d) {
      if (sumdep - maxmove > d) {
        fa[n - maxmove + 1] = nowpos;
        sumdep -= maxmove;
        twice++;
        maxmove -= 1;
        if (twice == 2) {
          if (rest.count(nowpos)) {
            nowpos = nowpos + 1;
            maxmove += 1;
            fa[n - maxmove + 2] = nowpos;
            maxmove = maxmove - 2;
            twice = 1;
          } else {
            nowpos += 1;
            twice = 0;
          }
        }
        continue;
      } else {
        int cut = sumdep - d;
        int togo = fa[n - cut];
        if (togo < n - maxmove + 1 && rest.count(togo) && twice == 0) {
          break;
        } else if (togo <= n - maxmove && togo + 1 <= n - maxmove &&
                   twice == 0) {
          fa[n - cut + 1] = togo + 1;
        } else {
          fa[n - cut + 1] = togo;
        }
        sumdep = d;
        break;
      }
    }
    if (sumdep != d)
      printf("NO\n");
    else {
      printf("YES\n");
      for (int i = 2; i <= n; i++) printf("%d ", fa[i]);
      puts("");
    }
  }
  return 0;
}