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 = 5e3 + 2;
int f[N], a[N];
int t, n, i, d, ncs, suc, m, p, tp;
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d%d", &n, &d);
    suc = 0;
    ncs = 1;
    a[1] = 1;
    tp = p = 1;
    m = 0;
    if (n * (n - 1) < d * 2) {
      puts("NO");
      continue;
    }
    for (i = 1; i <= n; i++) f[i] = i - 1;
    d = (n * (n - 1) >> 1) - d;
    for (i = n; i - 1 > ncs; i--) {
      if (i - 1 - ncs >= d) {
        f[i] = (i - 1 - d);
        suc = 1;
        break;
      }
      d -= (i - 1 - ncs);
      f[i] = a[++m];
      a[++p] = i;
      a[++p] = i;
      ++tp;
      if ((tp + 2 & tp + 1) == 0) {
        a[++p] = (++ncs);
        ++tp;
      }
    }
    if ((!suc) && (i - 1 - ncs >= d) && (d)) {
      f[i] = (i - 1 - d);
      suc = 1;
    }
    if (suc) {
      puts("YES");
      for (i = 2; i <= n; i++) printf("%d ", f[i]);
      puts("");
    } else
      puts("NO");
  }
}