#include <bits/stdc++.h>
using namespace std;
long long power(long long x, long long y, long long z) {
  long long ret = 1;
  while (y > 0) {
    if (y & 1) ret = (ret * x) % z;
    x = (x * x) % z;
    y >>= 1;
  }
  return ret;
}
const long long N = 1e4 + 5;
long long par[N], numChild[N];
vector<long long> levNodes[N];
int main() {
  long long t;
  scanf("%lld", &t);
  while (t--) {
    long long n, d;
    scanf("%lld%lld", &n, &d);
    long long mx = (n * (n - 1)) / 2LL;
    if (d > mx) {
      printf("NO\n");
      continue;
    }
    for (long long i = 0; i <= n - 1; ++i)
      numChild[i] = 1, par[i] = i - 1, levNodes[i] = {i};
    long long last = n - 1, sum = mx;
    while (sum > d) {
      numChild[last] = 0;
      long long curr = 0;
      while (curr + 1 < last && !levNodes[curr].size()) ++curr;
      if (curr + 1 == last) break;
      if (sum - d < last - curr - 1) {
        par[last] = last - 1 - (sum - d);
        sum = d;
        break;
      }
      long long node = levNodes[curr].back();
      par[last] = node;
      numChild[node]++;
      if (numChild[node] >= 2) levNodes[curr].pop_back();
      sum -= (last - curr - 1);
      levNodes[curr + 1].push_back(last);
      --last;
    }
    if (sum > d) {
      printf("NO\n");
    } else {
      printf("YES\n");
      for (long long i = 1; i <= n - 1; ++i) printf("%lld ", par[i] + 1);
      printf("\n");
    }
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
long long power(long long x, long long y, long long z) {
  long long ret = 1;
  while (y > 0) {
    if (y & 1) ret = (ret * x) % z;
    x = (x * x) % z;
    y >>= 1;
  }
  return ret;
}
const long long N = 1e4 + 5;
long long par[N], numChild[N];
vector<long long> levNodes[N];
int main() {
  long long t;
  scanf("%lld", &t);
  while (t--) {
    long long n, d;
    scanf("%lld%lld", &n, &d);
    long long mx = (n * (n - 1)) / 2LL;
    if (d > mx) {
      printf("NO\n");
      continue;
    }
    for (long long i = 0; i <= n - 1; ++i)
      numChild[i] = 1, par[i] = i - 1, levNodes[i] = {i};
    long long last = n - 1, sum = mx;
    while (sum > d) {
      numChild[last] = 0;
      long long curr = 0;
      while (curr + 1 < last && !levNodes[curr].size()) ++curr;
      if (curr + 1 == last) break;
      if (sum - d < last - curr - 1) {
        par[last] = last - 1 - (sum - d);
        sum = d;
        break;
      }
      long long node = levNodes[curr].back();
      par[last] = node;
      numChild[node]++;
      if (numChild[node] >= 2) levNodes[curr].pop_back();
      sum -= (last - curr - 1);
      levNodes[curr + 1].push_back(last);
      --last;
    }
    if (sum > d) {
      printf("NO\n");
    } else {
      printf("YES\n");
      for (long long i = 1; i <= n - 1; ++i) printf("%lld ", par[i] + 1);
      printf("\n");
    }
  }
  return 0;
}

```