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 = 1e5 + 5;
int t, n, d, pos[N], cnt[N], f[N], son[N];
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d%d", &n, &d);
    int mina = 0, maxa = n * (n - 1) / 2;
    int t = n, now = 1, dep = 0;
    while (t) {
      if (t >= now) {
        t -= now;
        mina += dep * now;
      } else {
        mina += dep * t;
        t = 0;
      }
      now *= 2;
      dep++;
    }
    if (d <= maxa && d >= mina)
      printf("YES\n");
    else {
      printf("NO\n");
      continue;
    }
    int delt = maxa - d;
    for (int i = 2; i <= n; i++) pos[i] = 1, cnt[i] = 1, son[i] = 2;
    cnt[1] = 1;
    son[1] = 2;
    int lw = n;
    while (delt) {
      int p = lw - 1;
      delt--;
      while (delt && pos[p - 1]) {
        p--;
        delt--;
      }
      pos[p]--;
      pos[p + 1] += 2;
      cnt[lw]--;
      cnt[p]++;
      lw--;
    }
    int tot = n - 1, pn = 2, sum = 1;
    now = 2;
    while (tot) {
      tot -= cnt[now];
      for (int i = pn; i <= pn + cnt[now] - 1; i++) {
        for (int j = sum; j > sum - cnt[now - 1]; j--)
          if (son[j]) {
            son[j]--;
            f[i] = j;
            break;
          }
      }
      sum += cnt[now];
      pn += cnt[now];
      now++;
    }
    for (int i = 2; i <= n; i++) printf("%d ", f[i]);
    printf("\n");
  }
  return 0;
}