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:

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<image>
#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-8;
const int maxn = 200000 + 5;
const int inf = 0x3f3f3f3f;
vector<int> vec[maxn];
int par[maxn];
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, d;
    cin >> n >> d;
    for (int i = 0; i <= n; i++) {
      vec[i].clear();
    }
    int cnt[n + 5];
    for (int i = 0; i < n; i++) {
      cnt[i] = 1;
    }
    int sum = (n * (n - 1)) / 2;
    int nowmin = 1;
    int nowmax = n - 1;
    if (sum < d) {
      cout << "NO" << endl;
      continue;
    }
    while (sum > d) {
      if (nowmin >= nowmax) break;
      if (sum - d >= nowmax - nowmin) {
        sum -= (nowmax - nowmin);
        cnt[nowmax]--;
        nowmax--;
        cnt[nowmin]++;
        if (cnt[nowmin] == (1 << nowmin)) {
          nowmin++;
        }
      } else {
        int x = sum - d;
        sum -= (sum - d);
        cnt[nowmax]--;
        nowmax--;
        cnt[nowmax - x + 1]++;
      }
    }
    if (sum > d) {
      cout << "NO" << endl;
      continue;
    }
    int cntt = 1;
    vec[0].push_back(1);
    for (int i = 1; i <= nowmax; i++) {
      for (int j = 1; j <= cnt[i]; j++) {
        vec[i].push_back(++cntt);
        par[cntt] = vec[i - 1][(j - 1) / 2];
      }
    }
    cout << "YES" << endl;
    for (int i = 2; i <= cntt; i++) {
      cout << par[i] << " ";
    }
    cout << endl;
  }
  return 0;
}
