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;
int cnt[5005], dep[5005], nex[5005];
int main() {
  int T;
  cin >> T;
  while (T--) {
    for (int i = 1; i <= 5000; i++) cnt[i] = nex[i] = 0;
    int n, d, mx = 1, now = 0, up = 0;
    cin >> n >> d;
    cnt[0] = 1;
    for (int i = 2; i <= n; i++) {
      if (cnt[mx] == 2 * cnt[mx - 1]) mx++;
      cnt[mx]++;
      now += mx;
      up += i - 1;
    }
    if (now > d || up < d) {
      puts("NO");
      continue;
    }
    puts("YES");
    while (now != d) {
      for (int i = mx; i >= 1; i--) {
        if (cnt[i + 1] + 1 <= 2 * (cnt[i] - 1)) {
          now++;
          cnt[i + 1]++;
          cnt[i]--;
          if (i == mx) mx++;
          break;
        }
      }
    }
    now = 0;
    for (int i = 1; i <= n; i++) {
      dep[i] = now;
      cnt[now]--;
      if (cnt[now] == 0) now++;
    }
    for (int i = 2; i <= n; i++) {
      for (int j = i - 1; j >= 1; j--) {
        if (dep[j] == dep[i] - 1 && nex[j] < 2) {
          nex[j]++;
          cout << j << ' ';
          break;
        }
      }
    }
    cout << endl;
  }
  return 0;
}
