#include <bits/stdc++.h>
using namespace std;
const int N = 5010;
int dep[N], fa[N];
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    int n, d;
    scanf("%d%d", &n, &d);
    dep[0] = -1;
    int pos, res = 0;
    for (int i = 1; i <= n; i++) {
      fa[i] = i >> 1;
      dep[i] = dep[i >> 1] + 1;
      res += dep[i];
      if (!(i & (i - 1))) pos = i;
    }
    d -= res;
    if (d < 0) {
      puts("NO");
      continue;
    }
    if (d == 0) {
      puts("YES");
      for (int i = 2; i <= n; i++) printf("%d ", fa[i]);
      puts("");
      continue;
    }
    for (int i = n; i; i--) {
      if ((i & (i - 1)) == 0) continue;
      d -= dep[pos] + 1 - dep[i];
      if (d <= 0) {
        while (d) pos = fa[pos], d++;
        d = 0;
        fa[i] = pos;
        break;
      }
      fa[i] = pos, dep[i] = dep[pos] + 1, pos = i;
    }
    if (d == 0) {
      puts("YES");
      for (int i = 2; i <= n; i++) printf("%d ", fa[i]);
      puts("");
      continue;
    } else
      puts("NO");
  }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 N = 5010;
int dep[N], fa[N];
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    int n, d;
    scanf("%d%d", &n, &d);
    dep[0] = -1;
    int pos, res = 0;
    for (int i = 1; i <= n; i++) {
      fa[i] = i >> 1;
      dep[i] = dep[i >> 1] + 1;
      res += dep[i];
      if (!(i & (i - 1))) pos = i;
    }
    d -= res;
    if (d < 0) {
      puts("NO");
      continue;
    }
    if (d == 0) {
      puts("YES");
      for (int i = 2; i <= n; i++) printf("%d ", fa[i]);
      puts("");
      continue;
    }
    for (int i = n; i; i--) {
      if ((i & (i - 1)) == 0) continue;
      d -= dep[pos] + 1 - dep[i];
      if (d <= 0) {
        while (d) pos = fa[pos], d++;
        d = 0;
        fa[i] = pos;
        break;
      }
      fa[i] = pos, dep[i] = dep[pos] + 1, pos = i;
    }
    if (d == 0) {
      puts("YES");
      for (int i = 2; i <= n; i++) printf("%d ", fa[i]);
      puts("");
      continue;
    } else
      puts("NO");
  }
}

```