#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int tc;
  cin >> tc;
  while (tc--) {
    int n, d;
    cin >> n >> d;
    vector<int> par(n), depth(n), deg(n);
    par[0] = -1, depth[0] = 0;
    for (int i = 1; i < n; i++) {
      par[i] = i - 1;
      depth[i] = i;
      deg[i - 1]++;
    }
    int cur = n * (n - 1) / 2;
    while (cur > d) {
      int idx = -1, dep = 1;
      for (int i = 0; i < n; i++) {
        if (depth[i] > dep) {
          idx = i;
          dep = depth[i];
        }
      }
      if (idx == -1) break;
      int j = -1, mn = 10000;
      for (int i = 0; i < n; i++) {
        if (depth[i] < dep - 1 && deg[i] < 2 &&
            cur - (dep - depth[i] - 1) >= d) {
          if (depth[i] < mn) {
            mn = depth[i];
            j = i;
          }
        }
      }
      if (j == -1) break;
      deg[j]++;
      deg[par[idx]]--;
      par[idx] = j;
      cur -= depth[idx] - 1 - mn;
      depth[idx] = mn + 1;
    }
    if (cur == d) {
      cout << "YES\n";
      for (int i = 1; i < n; i++) {
        cout << par[i] + 1 << " \n"[i == n - 1];
      }
    } else {
      cout << "NO\n";
    }
  }
}

### Prompt

Generate 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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int tc;
  cin >> tc;
  while (tc--) {
    int n, d;
    cin >> n >> d;
    vector<int> par(n), depth(n), deg(n);
    par[0] = -1, depth[0] = 0;
    for (int i = 1; i < n; i++) {
      par[i] = i - 1;
      depth[i] = i;
      deg[i - 1]++;
    }
    int cur = n * (n - 1) / 2;
    while (cur > d) {
      int idx = -1, dep = 1;
      for (int i = 0; i < n; i++) {
        if (depth[i] > dep) {
          idx = i;
          dep = depth[i];
        }
      }
      if (idx == -1) break;
      int j = -1, mn = 10000;
      for (int i = 0; i < n; i++) {
        if (depth[i] < dep - 1 && deg[i] < 2 &&
            cur - (dep - depth[i] - 1) >= d) {
          if (depth[i] < mn) {
            mn = depth[i];
            j = i;
          }
        }
      }
      if (j == -1) break;
      deg[j]++;
      deg[par[idx]]--;
      par[idx] = j;
      cur -= depth[idx] - 1 - mn;
      depth[idx] = mn + 1;
    }
    if (cur == d) {
      cout << "YES\n";
      for (int i = 1; i < n; i++) {
        cout << par[i] + 1 << " \n"[i == n - 1];
      }
    } else {
      cout << "NO\n";
    }
  }
}

```