#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int T;
  cin >> T;
  while (T--) {
    int n, d;
    cin >> n >> d;
    if (d > n * (n - 1) / 2) {
      cout << "NO" << '\n';
      continue;
    }
    int low = 0, D = 0;
    for (int i = 1; i <= n; ++i) {
      if ((i & (i - 1)) == 0) {
        ++D;
      }
      low += D - 1;
    }
    if (d < low) {
      cout << "NO" << '\n';
      continue;
    }
    vector<int> par;
    vector<int> cnt;
    vector<int> dep;
    vector<int> bad;
    par.resize(n);
    cnt.assign(n, 1);
    cnt[n - 1] = 0;
    bad.resize(n);
    dep.resize(n);
    iota(dep.begin(), dep.end(), 0);
    for (int i = 1; i < n; ++i) {
      par[i] = i - 1;
    }
    int cur = n * (n - 1) / 2;
    while (cur > d) {
      int u = -1;
      for (int i = 0; i < n; ++i) {
        if (bad[i] == 0 && cnt[i] == 0 && (u == -1 || dep[u] > dep[i])) {
          u = i;
        }
      }
      if (u == -1) {
        break;
      }
      int p = -1;
      for (int i = 0; i < n; ++i) {
        if (dep[u] - dep[i] == 2 && cnt[i] < 2) {
          p = i;
          break;
        }
      }
      if (p == -1) {
        bad[u] = 1;
        continue;
      }
      --cnt[par[u]];
      --dep[u];
      ++cnt[p];
      par[u] = p;
      --cur;
    }
    if (cur != d) {
      cout << "NO" << '\n';
    } else {
      cout << "YES" << '\n';
      for (int i = 1; i < n; ++i) {
        cout << par[i] + 1 << ' ';
      }
      cout << '\n';
    }
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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::sync_with_stdio(false);
  cin.tie(nullptr);
  int T;
  cin >> T;
  while (T--) {
    int n, d;
    cin >> n >> d;
    if (d > n * (n - 1) / 2) {
      cout << "NO" << '\n';
      continue;
    }
    int low = 0, D = 0;
    for (int i = 1; i <= n; ++i) {
      if ((i & (i - 1)) == 0) {
        ++D;
      }
      low += D - 1;
    }
    if (d < low) {
      cout << "NO" << '\n';
      continue;
    }
    vector<int> par;
    vector<int> cnt;
    vector<int> dep;
    vector<int> bad;
    par.resize(n);
    cnt.assign(n, 1);
    cnt[n - 1] = 0;
    bad.resize(n);
    dep.resize(n);
    iota(dep.begin(), dep.end(), 0);
    for (int i = 1; i < n; ++i) {
      par[i] = i - 1;
    }
    int cur = n * (n - 1) / 2;
    while (cur > d) {
      int u = -1;
      for (int i = 0; i < n; ++i) {
        if (bad[i] == 0 && cnt[i] == 0 && (u == -1 || dep[u] > dep[i])) {
          u = i;
        }
      }
      if (u == -1) {
        break;
      }
      int p = -1;
      for (int i = 0; i < n; ++i) {
        if (dep[u] - dep[i] == 2 && cnt[i] < 2) {
          p = i;
          break;
        }
      }
      if (p == -1) {
        bad[u] = 1;
        continue;
      }
      --cnt[par[u]];
      --dep[u];
      ++cnt[p];
      par[u] = p;
      --cur;
    }
    if (cur != d) {
      cout << "NO" << '\n';
    } else {
      cout << "YES" << '\n';
      for (int i = 1; i < n; ++i) {
        cout << par[i] + 1 << ' ';
      }
      cout << '\n';
    }
  }
  return 0;
}

```