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:

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<image>
#include <bits/stdc++.h>
using namespace std;
const int N = 5005;
int n, m, ans[N], T, dep[N], cnt[N];
void solve() {
  int sum = 0;
  for (int i = 1; i <= n; i++)
    ans[i] = i - 1, sum += i - 1, dep[i] = i, cnt[i] = 1;
  cnt[n] = 0;
  m -= sum;
  if (m > 0) {
    puts("NO");
    return;
  }
  for (int i = n; i >= 1; i--) {
    if (cnt[i]) continue;
    int num = i, pnt = 0;
    for (int j = 1; j <= n; j++)
      if (cnt[j] <= 1 && (dep[i] - dep[j] - 1) > pnt)
        pnt = dep[i] - dep[j] - 1, num = j;
    if (pnt > -m)
      for (int j = 1; j <= n; j++)
        if (cnt[j] <= 1 && dep[i] - dep[j] - 1 == -m && i != j) {
          num = j;
          break;
        }
    if (num == i) continue;
    cnt[ans[i]]--;
    cnt[num]++;
    ans[i] = num;
    m += dep[i] - dep[num] - 1;
    dep[i] = dep[num] + 1;
    if (m >= 0) break;
  }
  if (m != 0)
    puts("NO");
  else {
    puts("YES");
    for (int i = 2; i <= n; i++) cout << ans[i] << ' ';
    puts("");
  }
}
int main() {
  cin >> T;
  while (T--) {
    cin >> n >> m;
    solve();
  }
  return 0;
}
