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 m1[N];
int m2[N];
int ans[N];
void dfs(int n, int d, int p, int ind) {
  if (n == 0) {
    return;
  }
  ans[ind] = p;
  d -= (n - 1);
  for (int i = 0; i < n; i++) {
    if (m2[i] + m2[n - 1 - i] < d || m1[i] + m1[n - 1 - i] > d) {
      continue;
    }
    if (d - m1[i] > m2[n - 1 - i]) {
      dfs(i, d - m2[n - 1 - i], ind, ind + 1);
      dfs(n - 1 - i, m2[n - 1 - i], ind, ind + i + 1);
      break;
    } else {
      dfs(i, m1[i], ind, ind + 1);
      dfs(n - 1 - i, d - m1[i], ind, ind + i + 1);
      break;
    }
  }
}
int main() {
  m2[0] = 0;
  m1[0] = 0;
  int c = -1;
  for (int i = 1; i < N; i++) {
    m2[i] = m2[i - 1] + i - 1;
    int j = i;
    while (j % 2 == 0) {
      j /= 2;
    }
    if (j == 1) {
      c++;
    }
    m1[i] = m1[i - 1] + c;
  }
  int t;
  cin >> t;
  while (t--) {
    int n, d;
    cin >> n >> d;
    if (d > m2[n] || d < m1[n]) {
      cout << "NO" << endl;
      continue;
    }
    cout << "YES" << endl;
    dfs(n, d, -1, 0);
    for (int i = 1; i < n; i++) {
      cout << ans[i] + 1 << " ";
    }
    cout << endl;
  }
}