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 = 2e5 + 5;
int dep[N], fa[N], c[N], k[N];
signed main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n, d;
    scanf("%d %d", &n, &d);
    if (d > n * (n - 1) / 2) {
      puts("NO");
      continue;
    }
    int tmp = n - 1, mn = 0;
    for (int i = 1;; ++i) {
      if (tmp < (1 << i)) {
        mn += i * tmp;
        break;
      }
      tmp -= (1 << i), mn += i * (1 << i);
    }
    if (d < mn) {
      puts("NO");
      continue;
    }
    puts("YES");
    for (int i = 1; i <= n; ++i)
      fa[i] = i - 1, dep[i] = i - 1, c[i] = 1, k[i] = 0;
    c[n] = 0;
    int now = n * (n - 1) / 2;
    while (now > d) {
      int p, ok = 0, tmp;
      for (int i = 1; i <= n; ++i)
        if (!k[i] && !c[i]) {
          p = i;
          break;
        }
      for (int i = 1; i <= n; ++i)
        if (c[i] < 2 && dep[i] == dep[p] - 2) {
          tmp = i;
          ok = 1;
          break;
        }
      if (!ok) {
        k[p] = 1;
        continue;
      }
      --c[fa[p]], ++c[tmp], --dep[p], fa[p] = tmp, --now;
    }
    for (int i = 2; i <= n; ++i) printf("%d ", fa[i]);
    puts("");
  }
  return 0;
}