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 long long INF = (long long)1e9 + 5;
const long long MOD = (long long)1e9 + 7;
const int MAXN = 5005;
int mi[MAXN], mx[MAXN], last, pa[MAXN];
void preprocess() {
  int layer = 0, cur = 0, tot = 0;
  for (int i = 1; i < MAXN; i++) {
    mx[i] = i * (i - 1) / 2;
    if (cur < (1 << layer)) {
      cur++;
      tot += layer;
    } else {
      layer++;
      cur = 0;
      cur++;
      tot += layer;
    }
    mi[i] = tot;
  }
}
void solve(int sz, int tot, int p) {
  if (sz == 0) return;
  int v = last++;
  pa[v] = p;
  assert(mi[sz] <= tot && tot <= mx[sz]);
  int lson = sz - 1, rson = 0;
  while (mi[lson] + mi[rson] + sz - 1 > tot) {
    rson++;
    lson--;
  }
  tot -= sz - 1;
  for (int i = 0; i <= tot; i++) {
    if (mi[lson] <= i && i <= mx[lson] && mi[rson] <= tot - i &&
        tot - i <= mx[rson]) {
      solve(lson, i, v);
      solve(rson, tot - i, v);
      break;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  preprocess();
  int tc;
  cin >> tc;
  while (tc--) {
    int n, d;
    cin >> n >> d;
    if (d > mx[n] || d < mi[n]) {
      cout << "NO\n";
      continue;
    }
    last = 1;
    solve(n, d, 1);
    cout << "YES\n";
    for (int i = 2; i <= n; i++) {
      cout << pa[i] << " \n"[i == n];
    }
  }
  return 0;
}