Problem: 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>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const long long inf = 5e18;
void dfs(int i, int& minVertex, int& minDepth, int MIN_DEPTH_ALLOWED,
         int curr_d, vector<int> children[]) {
  if (children[i].size() < 2 and curr_d >= MIN_DEPTH_ALLOWED and
      minDepth > curr_d) {
    minVertex = i;
    minDepth = curr_d;
  }
  for (int j : children[i]) {
    dfs(j, minVertex, minDepth, MIN_DEPTH_ALLOWED, curr_d + 1, children);
  }
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, d;
    cin >> n >> d;
    vector<int> children[n + 1];
    int curr = (n * (n - 1)) / 2;
    int p[n + 1];
    for (int i = 1; i < n; i++) {
      children[i].push_back(i + 1);
      p[i + 1] = i;
    }
    int currVertex = n;
    bool flag = true;
    while (curr > d) {
      int minVertex = -1, minDepth = 1e5;
      int curr_d = currVertex - 1;
      dfs(1, minVertex, minDepth, curr_d - (curr - d) - 1, 0, children);
      if (minVertex == -1) {
        flag = false;
        break;
      }
      children[currVertex - 1].erase(find(children[currVertex - 1].begin(),
                                          children[currVertex - 1].end(),
                                          currVertex));
      children[minVertex].push_back(currVertex);
      p[currVertex] = minVertex;
      curr -= (curr_d - minDepth - 1);
      if (curr_d == minDepth) {
        break;
      }
      currVertex--;
    }
    if (curr != d) {
      flag = false;
    }
    if (!flag) {
      cout << "NO\n";
      continue;
    }
    cout << "YES\n";
    for (int i = 2; i <= n; i++) {
      cout << p[i] << ' ';
    }
    cout << endl;
  }
}