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;
using ull = unsigned long long;
using ll = long long;
using ld = long double;
const int mod = 1e9 + 7;
const int inf = INT_MAX;
const int N = 5000 + 5;
int p[N], mn[N], mx[N];
bool f(int n, int d) { return d >= mn[n] and d <= mx[n]; }
void construct(int idx, int par, int n, int d) {
  p[idx] = par;
  if (n == 1) return;
  if (f(n - 1, d - n + 1)) {
    construct(idx + 1, idx, n - 1, d - n + 1);
    return;
  }
  int target = d - n + 1;
  for (int i = 1; i < n; i++) {
    if (f(n - i - 1, target - mn[i])) {
      construct(idx + 1, idx, i, mn[i]);
      construct(idx + i + 1, idx, n - i - 1, target - mn[i]);
      return;
    }
    if (f(n - i - 1, target - mx[i])) {
      construct(idx + 1, idx, i, mx[i]);
      construct(idx + i + 1, idx, n - i - 1, target - mx[i]);
      return;
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  mn[1] = 0;
  for (int i = 2; i < N; i++) {
    mn[i] = mn[i - 1] + floor(log2(i));
    mx[i] = (i * i - i) / 2;
  }
  int t;
  cin >> t;
  while (t--) {
    int n, d;
    cin >> n >> d;
    if (!f(n, d)) {
      cout << "NO\n";
      continue;
    }
    construct(1, 0, n, d);
    cout << "YES\n";
    for (int i = 2; i <= n; i++) cout << p[i] << " \n"[i == n];
  }
  return 0;
}