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;
long long int gcd(long long int a, long long int b);
vector<string> split(string target, unsigned char c);
long long int n, k, t, m, x, y, len[5005], a[5005];
long long int ans;
long long int mod = 1e9 + 7;
bool isPrime(long long int h);
bool check() {
  if (n * (n - 1) / 2 < m) return 0;
  int num = n - 1, cnt = 1, sum = 0, mul = 2;
  while (num >= mul) {
    sum += cnt++ * mul;
    num -= mul;
    mul <<= 1;
  }
  sum += cnt * num;
  return m >= sum;
}
int main() {
  cout.precision(2);
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> t;
  while (t--) {
    cin >> n >> m;
    if (!check()) {
      cout << "NO\n";
      continue;
    }
    (memset(len, 0, sizeof(len)));
    ans = n - 1;
    for (int i = (1); i < (int)(n); i++) {
      int cnt = 0;
      for (int j = i + 1;
           j <= n && cnt < 2 && m <= len[i] * ans + ans * (ans + 1) / 2; j++) {
        if (!len[j]) {
          a[j] = i;
          cnt++;
          len[j] = len[i] + 1;
          ans--;
          m -= len[j];
        }
      }
    }
    cout << "YES\n";
    for (int i = 0; i < (int)(n - 1); i++) cout << a[i + 2] << " ";
    cout << "\n";
  }
}
long long int gcd(long long int a, long long int b) {
  return b == 0 ? a : gcd(b, a % b);
}
vector<string> split(string target, unsigned char c) {
  string d = "";
  vector<string> arr;
  for (auto n : target) {
    if (n != c)
      d += n;
    else if (d != "")
      arr.push_back(d), d = "";
  }
  if (d != "") arr.push_back(d), d = "";
  return arr;
}
bool isPrime(long long int g) {
  if ((g % 2 == 0 && g > 2) || g == 1) return 0;
  for (long long int i = 3; i * i <= g; i += 2)
    if (g % i == 0) return 0;
  return 1;
}