#include <bits/stdc++.h>
using namespace std;
const long long N = 5005;
long long T, n, d, mx[N], mn[N], f[N];
void dfs(long long tot, long long off, long long sum) {
  for (long long i = 0; i < tot; i++) {
    long long lsiz = i, rsiz = tot - 1 - i;
    long long vmin = mn[lsiz] + mn[rsiz] + tot - 1;
    long long vmax = mx[lsiz] + mx[rsiz] + tot - 1;
    if (vmin <= sum && sum <= vmax) {
      long long tmp = mn[lsiz];
      if (mn[lsiz] + mx[tot - 1 - lsiz] + tot - 1 < sum)
        tmp = sum - tot + 1 - mx[tot - 1 - lsiz];
      if (lsiz) {
        f[off + 1] = off;
        dfs(lsiz, off + 1, tmp);
      }
      if (rsiz) {
        f[off + 1 + lsiz] = off;
        dfs(rsiz, off + 1 + lsiz, sum - tot + 1 - tmp);
      }
      return;
    }
  }
}
signed main() {
  ios::sync_with_stdio(false);
  for (long long i = 1; i <= 5000; i++) {
    mx[i] = i * (i - 1) / 2;
    mn[i] = mn[i - 1] + (long long)(log2(i) + 1e-7);
  }
  cin >> T;
  while (T--) {
    cin >> n >> d;
    if (d < mn[n] || d > mx[n])
      cout << "NO" << endl;
    else {
      cout << "YES" << endl;
      dfs(n, 1, d);
      for (long long i = 2; i <= n; i++) cout << f[i] << " ";
      cout << endl;
    }
  }
}

### Prompt

Please create a solution in CPP 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 long long N = 5005;
long long T, n, d, mx[N], mn[N], f[N];
void dfs(long long tot, long long off, long long sum) {
  for (long long i = 0; i < tot; i++) {
    long long lsiz = i, rsiz = tot - 1 - i;
    long long vmin = mn[lsiz] + mn[rsiz] + tot - 1;
    long long vmax = mx[lsiz] + mx[rsiz] + tot - 1;
    if (vmin <= sum && sum <= vmax) {
      long long tmp = mn[lsiz];
      if (mn[lsiz] + mx[tot - 1 - lsiz] + tot - 1 < sum)
        tmp = sum - tot + 1 - mx[tot - 1 - lsiz];
      if (lsiz) {
        f[off + 1] = off;
        dfs(lsiz, off + 1, tmp);
      }
      if (rsiz) {
        f[off + 1 + lsiz] = off;
        dfs(rsiz, off + 1 + lsiz, sum - tot + 1 - tmp);
      }
      return;
    }
  }
}
signed main() {
  ios::sync_with_stdio(false);
  for (long long i = 1; i <= 5000; i++) {
    mx[i] = i * (i - 1) / 2;
    mn[i] = mn[i - 1] + (long long)(log2(i) + 1e-7);
  }
  cin >> T;
  while (T--) {
    cin >> n >> d;
    if (d < mn[n] || d > mx[n])
      cout << "NO" << endl;
    else {
      cout << "YES" << endl;
      dfs(n, 1, d);
      for (long long i = 2; i <= n; i++) cout << f[i] << " ";
      cout << endl;
    }
  }
}

```