#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n, d;
    scanf("%d%d", &n, &d);
    int mi = 0, mx = 0, tmp = n - 1, k = 2, max = 1;
    mx = n * (n - 1) / 2;
    int ans[5001] = {};
    while (tmp > 0) {
      ans[max] = min(k, tmp);
      mi += min(k, tmp) * max;
      tmp -= min(k, tmp);
      max++;
      k *= 2;
    }
    if (d > mx || d < mi) {
      printf("NO\n");
      continue;
    }
    tmp = max - 1;
    int sum = mi;
    while (sum != d) {
      if (ans[tmp] == 1) tmp--;
      ans[tmp]--;
      if (max - tmp >= d - sum) {
        ans[tmp + d - sum]++;
        sum = d;
      } else {
        sum += (max - tmp);
        ans[max]++;
        max++;
      }
    }
    printf("YES\n");
    vector<int> abc, xyz;
    tmp = 1;
    int curr = 2;
    abc.push_back(1);
    while (1) {
      if (ans[tmp] == 0) break;
      for (int i = 0; i < ans[tmp]; i++) {
        printf("%d ", abc[i / 2]);
        xyz.push_back(curr);
        curr++;
      }
      tmp++;
      swap(abc, xyz);
      xyz.clear();
    }
    printf("\n");
  }
}

### 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;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n, d;
    scanf("%d%d", &n, &d);
    int mi = 0, mx = 0, tmp = n - 1, k = 2, max = 1;
    mx = n * (n - 1) / 2;
    int ans[5001] = {};
    while (tmp > 0) {
      ans[max] = min(k, tmp);
      mi += min(k, tmp) * max;
      tmp -= min(k, tmp);
      max++;
      k *= 2;
    }
    if (d > mx || d < mi) {
      printf("NO\n");
      continue;
    }
    tmp = max - 1;
    int sum = mi;
    while (sum != d) {
      if (ans[tmp] == 1) tmp--;
      ans[tmp]--;
      if (max - tmp >= d - sum) {
        ans[tmp + d - sum]++;
        sum = d;
      } else {
        sum += (max - tmp);
        ans[max]++;
        max++;
      }
    }
    printf("YES\n");
    vector<int> abc, xyz;
    tmp = 1;
    int curr = 2;
    abc.push_back(1);
    while (1) {
      if (ans[tmp] == 0) break;
      for (int i = 0; i < ans[tmp]; i++) {
        printf("%d ", abc[i / 2]);
        xyz.push_back(curr);
        curr++;
      }
      tmp++;
      swap(abc, xyz);
      xyz.clear();
    }
    printf("\n");
  }
}

```