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 + 5;
int num[N], val[N][N], fa[N];
int n, m;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%d%d", &n, &m);
    if (n * (n - 1) / 2 < m) {
      printf("NO\n");
      continue;
    }
    m = n * (n - 1) / 2 - m;
    for (int i = 1; i <= n; i++) num[i] = 1, val[i][1] = i, fa[i] = i - 1;
    int now = 2;
    for (int i = n; m && i > now; i--) {
      int ret = min(m, i - now);
      m -= ret;
      val[i - ret][++num[i - ret]] = i;
      fa[i] = val[i - ret - 1][(num[i - ret] + 1) / 2];
      if (num[now] == 1 << now - 1) now++;
    }
    if (m)
      printf("NO\n");
    else {
      printf("YES\n");
      for (int i = 2; i <= n; i++) printf("%d%c", fa[i], " \n"[i == n]);
    }
  }
  return 0;
}