#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
using namespace std;
int T, N, D, cnt[5010], fl[5010];
queue<int> q[5010];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> T;
  while (T--) {
    memset(fl, 0, sizeof fl);
    cin >> N >> D;
    for (int i = 1; i <= N; ++i) {
      while (!q[i].empty()) q[i].pop();
    }
    int mx = 0;
    for (int i = 1; i < N; ++i) mx += i;
    int mn = 0, t = 4;
    for (int i = 2, d = 1; i <= N; ++i) {
      if (i == t) {
        ++d;
        t *= 2;
      }
      mn += d;
    }
    if (D < mn || D > mx) {
      cout << "NO\n";
      continue;
    }
    cout << "YES\n";
    iota(cnt + 2, cnt + N + 1, 1);
    int dif = mx - D, curd = 1, dt = 1, pos = 1, idx = N;
    while (dif) {
      int m = min(cnt[idx] - curd, dif);
      cnt[idx] -= m;
      dif -= m;
      dt--;
      --idx;
      if (dt == 0) {
        curd++;
        dt = (1 << curd) - 1;
      }
    }
    for (int i = 2; i <= N; ++i) q[cnt[i]].emplace(i);
    q[0].emplace(1);
    for (int i = 2; i <= N; ++i) {
      int v = q[cnt[i] - 1].front();
      cout << v << " ";
      if (fl[v])
        q[cnt[i] - 1].pop();
      else
        fl[v]++;
    }
    cout << '\n';
  }
}

### 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>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
using namespace std;
int T, N, D, cnt[5010], fl[5010];
queue<int> q[5010];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> T;
  while (T--) {
    memset(fl, 0, sizeof fl);
    cin >> N >> D;
    for (int i = 1; i <= N; ++i) {
      while (!q[i].empty()) q[i].pop();
    }
    int mx = 0;
    for (int i = 1; i < N; ++i) mx += i;
    int mn = 0, t = 4;
    for (int i = 2, d = 1; i <= N; ++i) {
      if (i == t) {
        ++d;
        t *= 2;
      }
      mn += d;
    }
    if (D < mn || D > mx) {
      cout << "NO\n";
      continue;
    }
    cout << "YES\n";
    iota(cnt + 2, cnt + N + 1, 1);
    int dif = mx - D, curd = 1, dt = 1, pos = 1, idx = N;
    while (dif) {
      int m = min(cnt[idx] - curd, dif);
      cnt[idx] -= m;
      dif -= m;
      dt--;
      --idx;
      if (dt == 0) {
        curd++;
        dt = (1 << curd) - 1;
      }
    }
    for (int i = 2; i <= N; ++i) q[cnt[i]].emplace(i);
    q[0].emplace(1);
    for (int i = 2; i <= N; ++i) {
      int v = q[cnt[i] - 1].front();
      cout << v << " ";
      if (fl[v])
        q[cnt[i] - 1].pop();
      else
        fl[v]++;
    }
    cout << '\n';
  }
}

```