#include <bits/stdc++.h>
using namespace std;
const int MaxN = 5005;
int test;
int n, d;
int p[MaxN], cnt[MaxN], bad[MaxN], depth[MaxN];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> test;
  while (test--) {
    cin >> n >> d;
    int k = n, t = 1, cd = 0, MaxLim = n * (n - 1) / 2, MinLim = 0;
    while (k > 0) {
      MinLim += min(k, t) * cd;
      cd++;
      k -= t;
      t *= 2;
    }
    if (d > MaxLim || d < MinLim) {
      cout << "NO\n";
      continue;
    }
    for (int i = 0; i < n; i++) {
      p[i] = i - 1;
      cnt[i] = 1;
      bad[i] = 0;
      depth[i] = i;
    }
    cnt[n - 1] = 0;
    int Count = MaxLim;
    while (Count > d) {
      int v = -1;
      for (int i = 0; i < n; i++) {
        if (bad[i] == 0 && cnt[i] == 0 && (v == -1 || depth[v] > depth[i])) {
          v = i;
        }
      }
      int Pd = -1;
      for (int i = 0; i < n; i++) {
        if (cnt[i] < 2 && depth[i] < depth[v] - 1 &&
            (Pd == -1 || depth[Pd] < depth[i])) {
          Pd = i;
        }
      }
      if (Pd == -1) {
        bad[v] = 1;
        continue;
      }
      cnt[p[v]]--;
      depth[v]--;
      cnt[Pd]++;
      p[v] = Pd;
      Count--;
    }
    cout << "YES\n";
    for (int i = 1; i < n; i++) {
      cout << p[i] + 1 << ' ';
    }
    cout << '\n';
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 MaxN = 5005;
int test;
int n, d;
int p[MaxN], cnt[MaxN], bad[MaxN], depth[MaxN];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> test;
  while (test--) {
    cin >> n >> d;
    int k = n, t = 1, cd = 0, MaxLim = n * (n - 1) / 2, MinLim = 0;
    while (k > 0) {
      MinLim += min(k, t) * cd;
      cd++;
      k -= t;
      t *= 2;
    }
    if (d > MaxLim || d < MinLim) {
      cout << "NO\n";
      continue;
    }
    for (int i = 0; i < n; i++) {
      p[i] = i - 1;
      cnt[i] = 1;
      bad[i] = 0;
      depth[i] = i;
    }
    cnt[n - 1] = 0;
    int Count = MaxLim;
    while (Count > d) {
      int v = -1;
      for (int i = 0; i < n; i++) {
        if (bad[i] == 0 && cnt[i] == 0 && (v == -1 || depth[v] > depth[i])) {
          v = i;
        }
      }
      int Pd = -1;
      for (int i = 0; i < n; i++) {
        if (cnt[i] < 2 && depth[i] < depth[v] - 1 &&
            (Pd == -1 || depth[Pd] < depth[i])) {
          Pd = i;
        }
      }
      if (Pd == -1) {
        bad[v] = 1;
        continue;
      }
      cnt[p[v]]--;
      depth[v]--;
      cnt[Pd]++;
      p[v] = Pd;
      Count--;
    }
    cout << "YES\n";
    for (int i = 1; i < n; i++) {
      cout << p[i] + 1 << ' ';
    }
    cout << '\n';
  }
  return 0;
}

```