#include <bits/stdc++.h>
using namespace std;
namespace RA {
int r(int p) { return 1ll * rand() * rand() % p; }
int r(int L, int R) { return r(R - L + 1) + L; }
}  // namespace RA
const int N = 5005;
int n, d;
vector<int> v[N];
int dep[N], p[N];
void go() {
  cin >> n >> d;
  for (int i = (0); i <= (N - 1); ++i) v[i].clear();
  for (int i = (1); i <= (n); ++i) dep[i] = 0;
  for (int i = (1); i <= (n); ++i) p[i] = 0;
  int cur = 0;
  for (int i = (1); i <= (n); ++i) {
    if (i > 1)
      dep[i] = dep[i / 2] + 1;
    else
      dep[i] = 0;
    d -= dep[i];
    cur = dep[i];
  }
  if (d < 0) return cout << "NO" << endl, void();
  for (int i = (1); i <= (n); ++i) v[dep[i]].push_back(i);
  for (int i = (n); i >= (0); --i)
    if (v[i].size() > 1) {
      if (d == 0) break;
      while (d > 0 && v[i].size() > 1) {
        if (d == 0) break;
        int x = v[i][1];
        v[i].erase(v[i].begin() + 1);
        if (cur + 1 - i <= d) {
          d -= cur + 1 - i;
          ++cur;
          v[cur].push_back(x);
        } else {
          v[i + d].push_back(x);
          d = 0;
        }
      }
    }
  if (d > 0) return cout << "NO" << endl, void();
  for (int i = (1); i <= (cur); ++i) {
    auto pu = v[i].begin();
    auto pf = v[i - 1].begin();
    int cnt = 0;
    while (pu != v[i].end()) {
      assert(pf != v[i - 1].end());
      p[*pu] = *pf;
      ++cnt;
      ++pu;
      if (cnt == 2) ++pf, cnt = 0;
    }
  }
  cout << "YES" << endl;
  for (int i = (2); i <= (n); ++i) cout << p[i] << " \n"[i == n];
}
int main() {
  int t;
  cin >> t;
  for (int i = (1); i <= (t); ++i) go();
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
namespace RA {
int r(int p) { return 1ll * rand() * rand() % p; }
int r(int L, int R) { return r(R - L + 1) + L; }
}  // namespace RA
const int N = 5005;
int n, d;
vector<int> v[N];
int dep[N], p[N];
void go() {
  cin >> n >> d;
  for (int i = (0); i <= (N - 1); ++i) v[i].clear();
  for (int i = (1); i <= (n); ++i) dep[i] = 0;
  for (int i = (1); i <= (n); ++i) p[i] = 0;
  int cur = 0;
  for (int i = (1); i <= (n); ++i) {
    if (i > 1)
      dep[i] = dep[i / 2] + 1;
    else
      dep[i] = 0;
    d -= dep[i];
    cur = dep[i];
  }
  if (d < 0) return cout << "NO" << endl, void();
  for (int i = (1); i <= (n); ++i) v[dep[i]].push_back(i);
  for (int i = (n); i >= (0); --i)
    if (v[i].size() > 1) {
      if (d == 0) break;
      while (d > 0 && v[i].size() > 1) {
        if (d == 0) break;
        int x = v[i][1];
        v[i].erase(v[i].begin() + 1);
        if (cur + 1 - i <= d) {
          d -= cur + 1 - i;
          ++cur;
          v[cur].push_back(x);
        } else {
          v[i + d].push_back(x);
          d = 0;
        }
      }
    }
  if (d > 0) return cout << "NO" << endl, void();
  for (int i = (1); i <= (cur); ++i) {
    auto pu = v[i].begin();
    auto pf = v[i - 1].begin();
    int cnt = 0;
    while (pu != v[i].end()) {
      assert(pf != v[i - 1].end());
      p[*pu] = *pf;
      ++cnt;
      ++pu;
      if (cnt == 2) ++pf, cnt = 0;
    }
  }
  cout << "YES" << endl;
  for (int i = (2); i <= (n); ++i) cout << p[i] << " \n"[i == n];
}
int main() {
  int t;
  cin >> t;
  for (int i = (1); i <= (t); ++i) go();
  return 0;
}

```