#include <bits/stdc++.h>
using namespace std;
mt19937 rng(time(0));
uniform_int_distribution<int> uid(-1e9, 1e9);
int p[5005], c[5005], u[5005], dx[5005];
void solve(int n, int d) {
  int x = 0, y = 0;
  for (int i = 1; i <= n; ++i) y += 1 - bool(i & (i - 1)), x += y - 1;
  if (!(x <= d && d <= n * (n - 1) / 2)) {
    cout << "NO\n";
    return;
  }
  fill(c, c + n, 1);
  fill(u, u + n, 0);
  iota(p + 1, p + n, 0);
  iota(dx, dx + n, 0);
  c[n - 1] = 0;
  x = n * (n - 1) / 2;
  while (x > d) {
    y = -1;
    for (int i = 0; i < n; ++i)
      if (!u[i] && c[i] == 0 && (y == -1 || dx[y] > dx[i])) y = i;
    int k = -1;
    for (int i = 0; i < n; ++i)
      if (c[i] < 2 && dx[i] < dx[y] - 1 && (k == -1 || dx[k] < dx[i])) k = i;
    if (k == -1)
      u[y] = 1;
    else
      --c[p[y]], --dx[y], ++c[k], p[y] = k, --x;
  }
  cout << "YES\n";
  for (int i = 1; i < n; ++i) cout << p[i] + 1 << ' ';
  cout << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n, d;
    cin >> n >> d;
    solve(n, d);
  }
}

### Prompt

Your task is to create 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;
mt19937 rng(time(0));
uniform_int_distribution<int> uid(-1e9, 1e9);
int p[5005], c[5005], u[5005], dx[5005];
void solve(int n, int d) {
  int x = 0, y = 0;
  for (int i = 1; i <= n; ++i) y += 1 - bool(i & (i - 1)), x += y - 1;
  if (!(x <= d && d <= n * (n - 1) / 2)) {
    cout << "NO\n";
    return;
  }
  fill(c, c + n, 1);
  fill(u, u + n, 0);
  iota(p + 1, p + n, 0);
  iota(dx, dx + n, 0);
  c[n - 1] = 0;
  x = n * (n - 1) / 2;
  while (x > d) {
    y = -1;
    for (int i = 0; i < n; ++i)
      if (!u[i] && c[i] == 0 && (y == -1 || dx[y] > dx[i])) y = i;
    int k = -1;
    for (int i = 0; i < n; ++i)
      if (c[i] < 2 && dx[i] < dx[y] - 1 && (k == -1 || dx[k] < dx[i])) k = i;
    if (k == -1)
      u[y] = 1;
    else
      --c[p[y]], --dx[y], ++c[k], p[y] = k, --x;
  }
  cout << "YES\n";
  for (int i = 1; i < n; ++i) cout << p[i] + 1 << ' ';
  cout << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n, d;
    cin >> n >> d;
    solve(n, d);
  }
}

```