#include <bits/stdc++.h>
using namespace std;
const int maxn = 5e3 + 5;
int n, d, p[maxn], fa[maxn];
vector<int> dep[maxn];
signed main() {
  int t;
  cin >> t;
  while (t--) {
    memset(p, 0, sizeof(p)), memset(fa, 0, sizeof(fa));
    cin >> n >> d;
    int sum = 0, bs = 0;
    for (int i = 1; i <= n;) {
      int rem = min(n - i + 1, 1 << bs);
      p[bs] = rem, sum += rem * bs;
      i += rem, bs++;
    }
    if (sum > d) {
      puts("no");
      continue;
    }
    int pos = bs;
    while (sum < d) {
      int k = 0;
      for (int i = 1; i <= n; i++)
        if ((p[i] - 1) * 2 >= p[i + 1] + 1) {
          k = i;
          break;
        }
      if (!k) break;
      sum++, p[k]--, p[k + 1]++;
    }
    if (sum < d) {
      puts("no");
      continue;
    }
    puts("yes");
    int cnt = 1;
    for (int i = 0; i <= n; i++) dep[i].clear();
    dep[0].push_back(1);
    dep[0].push_back(1);
    for (int i = 1; i <= n; i++) {
      while (p[i]--) {
        fa[++cnt] = dep[i - 1].back();
        dep[i].push_back(cnt);
        dep[i].push_back(cnt);
        dep[i - 1].pop_back();
      }
    }
    for (int i = 2; i <= n; i++) cout << fa[i] << ' ';
    cout << '\n';
  }
  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;
const int maxn = 5e3 + 5;
int n, d, p[maxn], fa[maxn];
vector<int> dep[maxn];
signed main() {
  int t;
  cin >> t;
  while (t--) {
    memset(p, 0, sizeof(p)), memset(fa, 0, sizeof(fa));
    cin >> n >> d;
    int sum = 0, bs = 0;
    for (int i = 1; i <= n;) {
      int rem = min(n - i + 1, 1 << bs);
      p[bs] = rem, sum += rem * bs;
      i += rem, bs++;
    }
    if (sum > d) {
      puts("no");
      continue;
    }
    int pos = bs;
    while (sum < d) {
      int k = 0;
      for (int i = 1; i <= n; i++)
        if ((p[i] - 1) * 2 >= p[i + 1] + 1) {
          k = i;
          break;
        }
      if (!k) break;
      sum++, p[k]--, p[k + 1]++;
    }
    if (sum < d) {
      puts("no");
      continue;
    }
    puts("yes");
    int cnt = 1;
    for (int i = 0; i <= n; i++) dep[i].clear();
    dep[0].push_back(1);
    dep[0].push_back(1);
    for (int i = 1; i <= n; i++) {
      while (p[i]--) {
        fa[++cnt] = dep[i - 1].back();
        dep[i].push_back(cnt);
        dep[i].push_back(cnt);
        dep[i - 1].pop_back();
      }
    }
    for (int i = 2; i <= n; i++) cout << fa[i] << ' ';
    cout << '\n';
  }
  return 0;
}

```