#include <bits/stdc++.h>
using namespace std;
int n, d, ans[5005], son[5005], dep[5005];
int main() {
  int T;
  cin >> T;
  while (T--) {
    scanf("%d%d", &n, &d);
    int log = 0, mn = 0;
    for (int i = 0; (1 << (i + 1)) - 1 <= n; ++i) {
      log = i;
      mn += (1 << i) * i;
    }
    mn += (n - (1 << (log + 1)) + 1) * (log + 1);
    int all = (0 + n - 1) * n / 2;
    if (mn > d || d > all) {
      printf("NO\n");
      continue;
    }
    printf("YES\n");
    dep[1] = 0;
    for (int i = 2; i <= n; ++i) {
      ans[i] = i - 1;
      son[i - 1] = 1;
      dep[i] = i - 1;
    }
    son[n] = 0;
    while (all > d) {
      int now = 1;
      for (int i = 1; i <= n; ++i) {
        if (son[i]) continue;
        if (dep[i] > dep[now]) now = i;
      }
      int mn = now;
      for (int i = 1; i <= n; ++i) {
        if (son[i] >= 2) continue;
        if (all - dep[now] + dep[i] + 1 >= d) {
          if (dep[i] < dep[mn]) mn = i;
        }
      }
      if (mn == now) break;
      son[mn]++;
      son[ans[now]]--;
      ans[now] = mn;
      all -= dep[now] - dep[mn] - 1;
      dep[now] = dep[mn] + 1;
    }
    for (int i = 2; i <= n; ++i) {
      printf("%d%s", ans[i], i == n ? "\n" : " ");
    }
  }
}

### Prompt

Your task is to create 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;
int n, d, ans[5005], son[5005], dep[5005];
int main() {
  int T;
  cin >> T;
  while (T--) {
    scanf("%d%d", &n, &d);
    int log = 0, mn = 0;
    for (int i = 0; (1 << (i + 1)) - 1 <= n; ++i) {
      log = i;
      mn += (1 << i) * i;
    }
    mn += (n - (1 << (log + 1)) + 1) * (log + 1);
    int all = (0 + n - 1) * n / 2;
    if (mn > d || d > all) {
      printf("NO\n");
      continue;
    }
    printf("YES\n");
    dep[1] = 0;
    for (int i = 2; i <= n; ++i) {
      ans[i] = i - 1;
      son[i - 1] = 1;
      dep[i] = i - 1;
    }
    son[n] = 0;
    while (all > d) {
      int now = 1;
      for (int i = 1; i <= n; ++i) {
        if (son[i]) continue;
        if (dep[i] > dep[now]) now = i;
      }
      int mn = now;
      for (int i = 1; i <= n; ++i) {
        if (son[i] >= 2) continue;
        if (all - dep[now] + dep[i] + 1 >= d) {
          if (dep[i] < dep[mn]) mn = i;
        }
      }
      if (mn == now) break;
      son[mn]++;
      son[ans[now]]--;
      ans[now] = mn;
      all -= dep[now] - dep[mn] - 1;
      dep[now] = dep[mn] + 1;
    }
    for (int i = 2; i <= n; ++i) {
      printf("%d%s", ans[i], i == n ? "\n" : " ");
    }
  }
}

```