#include <bits/stdc++.h>
using namespace std;
const int N = 5050;
int min_depth[N], max_depth[N];
void prec() {
  for (int n = 1; n <= 5000; n++) {
    int nn = n - 1, h = 0, cnt = 1;
    while (nn) {
      h++;
      cnt *= 2;
      min_depth[n] += min(nn, cnt) * h;
      nn -= min(nn, cnt);
    }
    max_depth[n] = n * (n - 1) / 2;
  }
}
int timer = 0;
vector<int> p;
void solver(int root, int n, int d) {
  if (n == 1) return;
  d -= (n - 1);
  for (int l = 0; l <= n - 1; l++) {
    int r = (n - 1) - l;
    if (!(min_depth[l] + min_depth[r] <= d && d <= max_depth[l] + max_depth[r]))
      continue;
    int flag = 0;
    for (int depth_l : {min_depth[l], max_depth[l]}) {
      if (min_depth[r] <= d - depth_l && d - depth_l <= max_depth[r]) {
        if (l) {
          int lv = timer++;
          p.push_back(root);
          solver(lv, l, depth_l);
        }
        if (r) {
          int rv = timer++;
          p.push_back(root);
          solver(rv, r, d - depth_l);
        }
        flag = 1;
        break;
      }
    }
    if (flag) break;
  }
}
void solve() {
  int n, d;
  cin >> n >> d;
  if (!(min_depth[n] <= d && d <= max_depth[n])) {
    cout << "NO\n";
    return;
  }
  cout << "YES\n";
  p.clear();
  timer = 2;
  solver(1, n, d);
  for (auto x : p) cout << x << " ";
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(NULL), cin.tie(0), cout.tie(0);
  cout.setf(ios::fixed), cout.precision(20);
  prec();
  int t;
  cin >> t;
  while (t--) solve();
}

### Prompt

Please formulate a CPP solution to the following 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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 5050;
int min_depth[N], max_depth[N];
void prec() {
  for (int n = 1; n <= 5000; n++) {
    int nn = n - 1, h = 0, cnt = 1;
    while (nn) {
      h++;
      cnt *= 2;
      min_depth[n] += min(nn, cnt) * h;
      nn -= min(nn, cnt);
    }
    max_depth[n] = n * (n - 1) / 2;
  }
}
int timer = 0;
vector<int> p;
void solver(int root, int n, int d) {
  if (n == 1) return;
  d -= (n - 1);
  for (int l = 0; l <= n - 1; l++) {
    int r = (n - 1) - l;
    if (!(min_depth[l] + min_depth[r] <= d && d <= max_depth[l] + max_depth[r]))
      continue;
    int flag = 0;
    for (int depth_l : {min_depth[l], max_depth[l]}) {
      if (min_depth[r] <= d - depth_l && d - depth_l <= max_depth[r]) {
        if (l) {
          int lv = timer++;
          p.push_back(root);
          solver(lv, l, depth_l);
        }
        if (r) {
          int rv = timer++;
          p.push_back(root);
          solver(rv, r, d - depth_l);
        }
        flag = 1;
        break;
      }
    }
    if (flag) break;
  }
}
void solve() {
  int n, d;
  cin >> n >> d;
  if (!(min_depth[n] <= d && d <= max_depth[n])) {
    cout << "NO\n";
    return;
  }
  cout << "YES\n";
  p.clear();
  timer = 2;
  solver(1, n, d);
  for (auto x : p) cout << x << " ";
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(NULL), cin.tie(0), cout.tie(0);
  cout.setf(ios::fixed), cout.precision(20);
  prec();
  int t;
  cin >> t;
  while (t--) solve();
}

```