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;
int getmn(int x, int n) {
  if (x * 2 >= n)
    return n;
  else
    return n + getmn(2 * x, n - 2 * x);
}
int getmx(int x, int n) { return n * (n + 1) / 2; }
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    int n, m;
    scanf("%d %d", &n, &m);
    if (getmn(1, n - 1) > m || getmx(1, n - 1) < m) {
      puts("NO");
      continue;
    }
    printf("YES\n");
    int st = 1;
    n--;
    int dep = 0;
    int all = 1;
    while (n) {
      int last = all - st + 1;
      st = min(n, 2 * st);
      dep++;
      while (st > 1 && n + getmx(st, n - (st)) < m) st--;
      for (int i = 1; i <= st; i += 2) {
        printf("%d ", last);
        if (i + 1 <= st) printf("%d ", last);
        last++;
      }
      m -= n;
      all += st;
      n -= st;
    }
    printf("\n");
  }
  return 0;
}