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;
int f[5010];
int deep[5010];
int a[5010];
int dd[5010];
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    int n, d;
    scanf("%d%d", &n, &d);
    int ans = 0;
    for (int i = 2; i <= n; i++) {
      f[i] = i / 2;
      deep[i] = deep[f[i]] + 1;
      ans += deep[i];
    }
    vector<int> vec;
    if (d < ans || d > n * (n - 1) / 2) {
      printf("NO\n");
      continue;
    } else {
      printf("YES\n");
      for (int i = 1; i <= n - 1; i++) dd[i] = 1;
      for (int i = 2; i <= n; i++) a[i] = i - 1;
      int sum = n * (n - 1) / 2;
      map<int, int> mp;
      mp[1] = 1;
      int dd = 1;
      for (int i = n; i >= 2; i--) {
        if (sum - (i - 1 - dd) >= d) {
          sum -= (i - 1 - dd);
          map<int, int>::iterator it;
          it = mp.begin();
          int x = it->first;
          a[i] = x;
          mp[x]++;
          vec.push_back(i);
          if (mp[x] == 2) mp.erase(x);
          if (mp.empty()) {
            mp[dd + 1] = 1;
            for (int k = 0; k < vec.size(); k++) mp[vec[k]] = 0;
            vec.clear();
            dd++;
          }
          if (sum == d) break;
        } else {
          a[i] = i - 1 - (sum - d);
          break;
        }
      }
      for (int i = 2; i <= n; i++) {
        if (i != 2) printf(" ");
        printf("%d", a[i]);
      }
      printf("\n");
    }
  }
  return 0;
}
