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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#include <bits/stdc++.h>
using namespace std;
const int N = 5007, oo = 1e9;
int madp[N], midp[N];
int parent[N];
int v;
void dfs(int num, int desir, int p) {
  assert((midp[num] <= desir && desir <= madp[num]));
  if (num == 0) return;
  int th = v;
  parent[v++] = p;
  if (num == 1) return;
  desir -= num - 1;
  for (int d = 0; d < num; d++) {
    int dd = num - d - 1;
    if ((midp[dd] <= desir - midp[d] && desir - midp[d] <= madp[dd])) {
      dfs(d, midp[d], th);
      dfs(dd, desir - midp[d], th);
      break;
    }
    if ((midp[dd] <= desir - madp[d] && desir - madp[d] <= madp[dd])) {
      dfs(d, madp[d], th);
      dfs(dd, desir - madp[d], th);
      break;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  for (int i = 0; i < N; i++) midp[i] = oo, madp[i] = -1;
  for (int i = 0; i < 2; i++) midp[i] = madp[i] = 0;
  for (int i = 2; i < N; i++) {
    for (int d = 0; d < i; d++) {
      int dd = i - d - 1;
      midp[i] = min(midp[i], midp[d] + midp[dd] + i - 1);
      madp[i] = max(madp[i], madp[d] + madp[dd] + i - 1);
    }
  }
  int t;
  cin >> t;
  stringstream ss;
  while (t--) {
    int n, d;
    cin >> n >> d;
    if ((midp[n] <= d && d <= madp[n])) {
      v = 0;
      dfs(n, d, -1);
      ss << "YES" << endl;
      for (int i = 1; i < n; i++) ss << parent[i] + 1 << " \n"[i == n - 1];
    } else
      ss << "NO" << endl;
  }
  cout << ss.str();
  return 0;
}
