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 = 2e5 + 10;
int t, n, d;
int f[N], de[N], L[N], R[N], fa[N];
int main() {
  ios::sync_with_stdio(false);
  for (cin >> t; t; t--) {
    cin >> n >> d;
    int sta = 0;
    for (int i = 1; i <= n; i++) {
      f[i] = 0;
      L[i] = R[i] = 0;
      de[i] = 0;
      fa[i] = 0;
    }
    for (int i = 2; i <= n; i++)
      f[i] = i - 1, de[i] = i - 1, L[i - 1] = i, fa[i] = i - 1;
    int now = (1 + n - 1) * (n - 1) / 2;
    for (int i = n; i >= 2 && now > d; i--) {
      int dis = now - d, res = -1, pos;
      for (int j = 1; j <= n; j++) {
        if (j == i) continue;
        if (L[j] && R[j]) {
          continue;
        }
        if (de[j] + 1 >= de[i]) continue;
        if (de[i] - de[j] - 1 > dis) continue;
        if (de[i] - 1 - de[j] > res) {
          res = de[i] - de[j] - 1;
          pos = j;
        }
      }
      if (res == -1) {
        sta = 1;
        break;
      }
      now -= res;
      if (pos == 1) {
        if (L[fa[i]] == i) {
          L[fa[i]] = 0;
        } else
          R[fa[i]] = 0;
        R[pos] = i;
        L[i] = R[i] = 0;
        f[i] = pos;
        fa[i] = pos;
        de[i] = 1;
      } else {
        if (L[fa[i]] == i) {
          L[fa[i]] = 0;
        } else
          R[fa[i]] = 0;
        fa[i] = pos;
        L[i] = R[i] = 0;
        if (!L[pos])
          L[pos] = i;
        else
          R[pos] = i;
        f[i] = pos;
        de[i] = de[pos] + 1;
      }
    }
    if (sta || now != d) {
      cout << "NO\n";
      continue;
    }
    cout << "YES\n";
    for (int i = 2; i <= n; i++) cout << f[i] << ' ';
    cout << '\n';
  }
  return 0;
}