#include <bits/stdc++.h>
using namespace std;
constexpr int MOD = 998244353;
constexpr long double EPS = 1e-12;
long long int N, M, K, H, W, L, R;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> K;
  while (K--) {
    cin >> N >> M;
    L = 0;
    R = N * (N - 1) / 2;
    for (int i = 1; i < N; i++) {
      int index = i + 1;
      while (index) {
        L++;
        index >>= 1;
      }
      L--;
    }
    if (L > M || R < M) {
      cout << "NO\n";
      continue;
    }
    cout << "YES\n";
    vector<vector<int>> h(N + 1);
    for (int i = 0; i < N; i++) {
      int index = i + 1;
      int d = -1;
      while (index) {
        index >>= 1;
        d++;
      }
      h[d].push_back(i);
    }
    while (L != M) {
      for (int i = 1; i < N; i++) {
        if (h[i].size() * 2 >= h[i + 1].size() + 3) {
          h[i + 1].push_back(h[i].back());
          h[i].pop_back();
          L++;
          break;
        }
      }
    }
    vector<int> p(N);
    for (int i = 1; i < N; i++) {
      for (int j = 0; j < h[i].size(); j++) {
        p[h[i][j]] = h[i - 1][j / 2];
      }
    }
    for (int i = 1; i < N; i++) cout << p[i] + 1 << " ";
    cout << endl;
  }
  return 0;
}

### 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;
constexpr int MOD = 998244353;
constexpr long double EPS = 1e-12;
long long int N, M, K, H, W, L, R;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> K;
  while (K--) {
    cin >> N >> M;
    L = 0;
    R = N * (N - 1) / 2;
    for (int i = 1; i < N; i++) {
      int index = i + 1;
      while (index) {
        L++;
        index >>= 1;
      }
      L--;
    }
    if (L > M || R < M) {
      cout << "NO\n";
      continue;
    }
    cout << "YES\n";
    vector<vector<int>> h(N + 1);
    for (int i = 0; i < N; i++) {
      int index = i + 1;
      int d = -1;
      while (index) {
        index >>= 1;
        d++;
      }
      h[d].push_back(i);
    }
    while (L != M) {
      for (int i = 1; i < N; i++) {
        if (h[i].size() * 2 >= h[i + 1].size() + 3) {
          h[i + 1].push_back(h[i].back());
          h[i].pop_back();
          L++;
          break;
        }
      }
    }
    vector<int> p(N);
    for (int i = 1; i < N; i++) {
      for (int j = 0; j < h[i].size(); j++) {
        p[h[i][j]] = h[i - 1][j / 2];
      }
    }
    for (int i = 1; i < N; i++) cout << p[i] + 1 << " ";
    cout << endl;
  }
  return 0;
}

```