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>
#include <bits/stdc++.h>
using namespace std;
vector<int> v[2005];
int par[2005];
int n, d;
void solve() {
  scanf("%d%d", &n, &d);
  int rem = n;
  int cur = 0;
  bool f = true;
  if (d > (n * (n - 1) / 2)) {
    printf("NO\n");
    return;
  }
  for (int i = 0; i < n && rem; i++) {
    v[i].clear();
    if (cur) par[cur] = v[i - 1][0];
    d -= i;
    rem--;
    v[i].push_back(cur);
    cur++;
    if (d < 0) {
      f = false;
      break;
    }
    for (int j = 1; j < (1 << i) && rem; j++) {
      if (d < 0) {
        f = false;
        break;
      }
      if (d - (i + ((rem - 2) * (rem - 1)) / 2 + (i + 1) * (rem - 1)) > 0)
        break;
      par[cur] = v[i - 1][j / 2];
      d -= i;
      rem--;
      v[i].push_back(cur);
      cur++;
    }
  }
  if (f && d == 0) {
    printf("YES\n");
    for (int i = 1; i < n; i++) printf("%d ", par[i] + 1);
    printf("\n");
  } else
    printf("NO\n");
}
int main() {
  int t = 1;
  scanf("%d", &t);
  while (t--) solve();
  return 0;
}
