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>
#pragma GCC optimize(3)
using namespace std;
inline int read() {
  char c = getchar();
  int x = 0;
  bool f = 0;
  for (; !isdigit(c); c = getchar()) f ^= !(c ^ 45);
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
  if (f) x = -x;
  return x;
}
int n, k, sum, maxd;
int dep[5005], fa[5005];
deque<int> s[5005];
inline void push(int u, int fa) {
  dep[u] = dep[fa] + 1;
  s[dep[u]].push_back(u);
  sum += dep[u], maxd = max(maxd, dep[u]);
}
void work() {
  for (register int i = (1); i <= (5000); ++i) s[i].clear();
  n = read(), k = read();
  maxd = sum = 0;
  s[0].push_back(1);
  for (register int i = (1); i <= (n); ++i) {
    int ls = i << 1, rs = ls | 1;
    if (ls <= n) push(ls, i);
    if (rs <= n) push(rs, i);
  }
  if (sum > k || k > n * (n - 1) / 2) {
    puts("NO");
    return;
  }
  puts("YES");
  while (sum < k) {
    for (register int i = (maxd); i >= (0); --i)
      if ((s[i].size()) > 1 && (s[i + 1].size()) < ((s[i].size()) - 1) * 2) {
        s[i + 1].push_back(s[i].back());
        s[i].pop_back();
        maxd = max(maxd, i + 1);
        sum++;
        break;
      }
  }
  for (register int i = (0); i <= (maxd - 1); ++i) {
    for (register int j = (1); j <= ((s[i].size())); ++j) {
      int u = s[i][j - 1];
      int ls = (j - 1) * 2, rs = ls + 1;
      if (ls < (s[i + 1].size())) fa[s[i + 1][ls]] = u;
      if (rs < (s[i + 1].size())) fa[s[i + 1][rs]] = u;
    }
  }
  for (register int i = (2); i <= (n); ++i) cout << fa[i] << ' ';
  puts("");
}
signed main() {
  int T = read();
  while (T--) work();
  return 0;
}