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 long long N = 5e3 + 5;
long long t, n, d, fa[N], dep[N], a[N], mx, ans;
bool vis[N];
signed main() {
  scanf("%lld", &t);
  while (t--) {
    scanf("%lld%lld", &n, &d);
    memset(vis, 0, sizeof(vis)), mx = 0, a[0] = 1;
    for (long long i = 2; i <= n; i++)
      fa[i] = i / 2, dep[i] = dep[fa[i]] + 1, d -= dep[i], mx = max(mx, dep[i]);
    if (d < 0) {
      puts("NO");
      continue;
    }
    long long x = n;
    while (x) a[dep[x]] = x, vis[x] = 1, x = fa[x];
    for (long long i = n; i >= 1; i--) {
      if (vis[i]) continue;
      long long pre = mx;
      while (dep[fa[i]] < pre && d) {
        fa[i] = a[dep[fa[i]] + 1], dep[i] = dep[fa[i]] + 1;
        if (dep[i] > mx) mx++, a[mx] = i, vis[i] = 1;
        d--;
      }
    }
    if (d) {
      puts("NO");
      continue;
    }
    puts("YES");
    for (long long i = 2; i <= n; i++)
      printf("%lld%c", fa[i], i == n ? '\n' : ' ');
  }
  return 0;
}