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;
bool check(long long n, long long pos) { return n & (1LL << pos); }
long long Set(long long n, long long pos) { return n = n | (1LL << pos); }
long long child[5003];
long long par[5003];
long long lev[5003];
bool chk(long long n, long long m) {
  long long ub = (n * (n - 1)) / 2;
  long long lb = 0;
  long long cnt = 2;
  long long sum = 0;
  for (long long i = 1; i <= n; i++) {
    if (i == cnt) {
      cnt *= 2;
      sum += 1;
    }
    lb += sum;
  }
  if (lb <= m && m <= ub) return 0;
  return 1;
}
int main() {
  long long i, j, k, l, m, n, o, r, q;
  long long testcase;
  long long input, flag, tag, ans;
  scanf("%lld", &testcase);
  for (long long cs = 1; cs <= testcase; cs++) {
    scanf("%lld", &n);
    scanf("%lld", &m);
    if (chk(n, m)) {
      cout << "NO" << endl;
      continue;
    }
    cout << "YES" << endl;
    for (i = 1; i <= n; i++) par[i] = i - 1;
    for (i = 1; i <= n; i++) {
      lev[i] = i - 1;
      if (i < n)
        child[i] = 1;
      else
        child[i] = 0;
    }
    long long cnt = (n * (n - 1)) / 2;
    long long last = n;
    while (cnt > m) {
      long long levl = lev[last];
      bool on = 1;
      for (i = 1; i <= n; i++) {
        if (lev[i] + 2 == levl && child[i] < 2) {
          child[par[last]]--;
          par[last] = i;
          child[i]++;
          lev[last]--;
          on = 0;
          break;
        }
      }
      if (on)
        last--;
      else
        cnt--;
    }
    for (i = 2; i <= n; i++) cout << par[i] << " ";
    cout << endl;
  }
}