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 = 5010;
int n, d, p[N], mn[N], mx[N];
int cnt = 1;
void DFS(int prnt, int cur, int d) {
  if (cur < 1) return;
  int node = cnt++;
  p[node] = prnt;
  if (cur == 1) return;
  cur--;
  d -= cur;
  for (int i = 0; i <= cur; i++) {
    if (d >= mn[i] + mn[cur - i] && d <= mx[i] + mx[cur - i]) {
      for (int k = 0; k <= d; k++) {
        if (k >= mn[i] && k <= mx[i] && (d - k) >= mn[cur - i] &&
            (d - k) <= mx[cur - i]) {
          DFS(node, i, k);
          DFS(node, cur - i, d - k);
          return;
        }
      }
    }
  }
}
void solve() {
  scanf("%d%d", &n, &d);
  cnt = 1;
  if (d < mn[n] || d > mx[n]) {
    puts("NO");
    return;
  }
  puts("YES");
  cnt = 1;
  DFS(-1, n, d);
  for (int i = 2; i <= n; i++) {
    if (i > 2) putchar(' ');
    printf("%d", p[i]);
  }
  puts("");
}
int main() {
  mn[1] = mx[1] = mn[0] = mx[0] = 0;
  for (int i = 2; i < N; i++) {
    mn[i] = 1000000010;
    mx[i] = -1000000010;
    for (int j = 0; j < i; j++) {
      mn[i] = min(mn[i], i - 1 + mn[j] + mn[i - j - 1]);
      mx[i] = max(mx[i], i - 1 + mx[j] + mx[i - j - 1]);
    }
  }
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}