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;
const int maxn = 5e3 + 50;
struct TreeNode {
  int parent, left, right, depth;
} tnode[maxn];
int tot, n, d, vis[maxn];
void init() {
  for (int i = 1; i <= tot; ++i)
    tnode[i].parent = tnode[i].left = tnode[i].right = vis[i] = 0;
  tot = n;
  tnode[1].depth = 0;
  for (int i = 2; i <= n; ++i) {
    tnode[i].parent = i - 1;
    tnode[i - 1].left = i;
    tnode[i].depth = tnode[i - 1].depth + 1;
  }
  tnode[1].parent = -1;
}
void change(int son, int prt) {
  int curp = tnode[son].parent;
  if (tnode[curp].left == son)
    tnode[curp].left = 0;
  else
    tnode[curp].right = 0;
  tnode[son].parent = prt;
  if (!tnode[prt].left)
    tnode[prt].left = son;
  else
    tnode[prt].right = son;
  tnode[son].depth = tnode[prt].depth + 1;
}
int main() {
  int t, upThreshold, downThreshold, i, j;
  ios::sync_with_stdio(false);
  cin >> t;
  while (t--) {
    cin >> n >> d;
    upThreshold = n * (n - 1) / 2;
    int cnt = 0;
    downThreshold = 0;
    for (i = 2; i <= n; ++i) {
      if (!(i & (i - 1))) ++cnt;
      downThreshold += cnt;
    }
    if (d > upThreshold || d < downThreshold) {
      cout << "NO\n";
      continue;
    }
    init();
    int cur = upThreshold;
    int flag = 0;
    while (cur > d) {
      for (i = 1; i <= tot; ++i) {
        if (vis[i]) continue;
        if (!tnode[i].left && !tnode[i].right) break;
      }
      if (i > tot) {
        flag = 1;
        break;
      }
      for (j = 1; j <= tot; ++j) {
        if (tnode[j].depth == tnode[i].depth - 2 &&
            (!tnode[j].left || !tnode[j].right))
          break;
      }
      if (j < tot) {
        --cur;
        change(i, j);
      } else {
        vis[i] = 1;
      }
    }
    if (flag)
      cout << "NO\n";
    else {
      cout << "YES\n";
      for (i = 2; i < n; ++i) cout << tnode[i].parent << ' ';
      cout << tnode[i].parent << '\n';
    }
  }
  return 0;
}
