#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
int f[N], depth[N], node[N], flag[N];
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n, d;
    cin >> n >> d;
    for (int i = 2; i <= n; i++) {
      f[i] = i / 2;
      depth[i] = depth[f[i]] + 1;
      d -= depth[i];
    }
    int maxn = depth[n];
    if (d < 0) {
      cout << "NO\n";
      continue;
    }
    int now = n;
    for (int i = 1; i <= n; i++) flag[i] = 0;
    while (now) {
      flag[now] = 1;
      node[depth[now]] = now;
      now = f[now];
    }
    for (int i = n; i > 1; i--) {
      if (flag[i]) continue;
      int pre = maxn;
      while (d && depth[i] - 1 < pre) {
        f[i] = node[depth[i]];
        depth[i]++;
        if (depth[i] > maxn) {
          maxn = depth[i];
          flag[i] = 1;
          node[depth[i]] = i;
        }
        d--;
      }
    }
    if (d)
      cout << "NO\n";
    else {
      cout << "YES\n";
      for (int i = 2; i <= n; i++) cout << f[i] << " ";
      cout << endl;
    }
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 = 2e5 + 10;
int f[N], depth[N], node[N], flag[N];
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n, d;
    cin >> n >> d;
    for (int i = 2; i <= n; i++) {
      f[i] = i / 2;
      depth[i] = depth[f[i]] + 1;
      d -= depth[i];
    }
    int maxn = depth[n];
    if (d < 0) {
      cout << "NO\n";
      continue;
    }
    int now = n;
    for (int i = 1; i <= n; i++) flag[i] = 0;
    while (now) {
      flag[now] = 1;
      node[depth[now]] = now;
      now = f[now];
    }
    for (int i = n; i > 1; i--) {
      if (flag[i]) continue;
      int pre = maxn;
      while (d && depth[i] - 1 < pre) {
        f[i] = node[depth[i]];
        depth[i]++;
        if (depth[i] > maxn) {
          maxn = depth[i];
          flag[i] = 1;
          node[depth[i]] = i;
        }
        d--;
      }
    }
    if (d)
      cout << "NO\n";
    else {
      cout << "YES\n";
      for (int i = 2; i <= n; i++) cout << f[i] << " ";
      cout << endl;
    }
  }
  return 0;
}

```