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 int N = 5 * 1000 + 100;
int n, d, par[N], deg[N], h[N];
unordered_set<int> s[N][3];
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int Q;
  cin >> Q;
  while (Q--) {
    cin >> n >> d;
    int low = 0, high = n * (n - 1) / 2;
    h[1] = 0, deg[1] = 0, par[1] = -1;
    for (int i = 2; i <= n; i++)
      par[i] = i / 2, deg[i / 2]++, h[i] = h[i / 2] + 1;
    int H = 0;
    for (int i = 1; i <= n; i++)
      low += h[i], s[h[i]][deg[i]].insert(i), H = max(H, h[i]);
    if (d < low || d > high)
      cout << "NO\n";
    else {
      cout << "YES\n";
      while (low < d) {
        for (int i = 1; i < n && low < d; i++) {
          int sz[3] = {(int)(s[i][0].size()), (int)(s[i][1].size()),
                       (int)(s[i][2].size())},
              v = -1, u = -1;
          if (sz[0] >= 1) v = *s[i][0].begin();
          if (sz[0] >= 2)
            u = *++s[i][0].begin();
          else if (sz[1] >= 1)
            u = *s[i][1].begin();
          if (u != -1 && v != -1) {
            s[i - 1][deg[par[v]]].erase(par[v]);
            deg[par[v]]--;
            s[i - 1][deg[par[v]]].insert(par[v]);
            s[i][0].erase(v);
            par[v] = u, h[v]++;
            s[i][deg[u]].erase(u);
            deg[u]++;
            s[i][deg[u]].insert(u);
            s[i][0].erase(v);
            s[i + 1][0].insert(v);
            low++;
          }
        }
      }
      for (int i = 2; i <= n; i++) cout << par[i] << ' ';
      cout << '\n';
    }
    for (int i = 0; i <= n; i++)
      deg[i] = par[i] = h[i] = 0, s[i][0].clear(), s[i][1].clear(),
      s[i][2].clear();
  }
  return 0;
}
