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;
using ll = long long;
using db = double;
using pii = pair<int, int>;
using vi = vector<int>;
int _, n, d, num[5100];
signed main() {
  for (scanf("%d", &_); _; _--) {
    scanf("%d %d", &n, &d);
    int dmin = 0, dmax = 0, h = 0, tmp = n;
    memset(num, 0, sizeof num);
    while (tmp) {
      h++;
      if (tmp >= 1 << h - 1) {
        tmp -= 1 << h - 1;
        dmin += (1 << h - 1) * (h - 1);
        num[h] = 1 << h - 1;
      } else {
        dmin += tmp * (h - 1);
        num[h] = tmp;
        tmp = 0;
      }
    }
    for (int i = 1; i <= n; i++) dmax += i - 1;
    if (d < dmin || d > dmax) {
      puts("NO");
      continue;
    }
    d -= dmin;
    while (d) {
      d--;
      for (int i = h + 1; i >= 1; i--) {
        if (num[i] + 1 <= 2 * (num[i - 1] - 1)) {
          num[i]++, num[i - 1]--;
          if (i == h + 1) h++;
          break;
        }
      }
    }
    puts("YES");
    int now = 1;
    for (int i = 2; i <= h; i++) {
      for (int j = 1; j <= num[i]; j++) printf("%d ", now + (j + 1) / 2 - 1);
      now += num[i - 1];
    }
    puts("");
  }
  return 0;
}