#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 1;
const int NN = 1e3 + 1;
const long long MOD = 1e9 + 7;
const long long oo = 1e18 + 7;
const int BASE = 10000;
void query() {
  int n, D;
  cin >> n >> D;
  vector<int> p(n + 1), c(n + 1), d(n + 1);
  vector<set<int>> de(n);
  for (int i = 1; i <= n; i++) {
    p[i] = i - 1;
    c[i] = (i != n ? 1 : 0);
    d[i] = i - 1;
    de[i - 1].insert(i);
  }
  int cur = (n - 1) * n / 2;
  int bot = 0, m = n, s = 1, cnt = 0;
  while (m) {
    bot += min(m, s) * cnt;
    m -= min(m, s);
    s *= 2;
    cnt++;
  }
  if (D > cur || D < bot) {
    cout << "NO"
         << "\n";
    return;
  }
  while (cur > D) {
    int found = 0;
    for (int i = 1; i <= n; i++) {
      if (!c[i] && d[i] >= 2 && de[d[i] - 2].size()) {
        found = 1;
        c[p[i]]--;
        if (c[p[i]] == 1) {
          de[d[p[i]]].insert(p[i]);
        }
        int t = *de[d[i] - 2].begin();
        c[t]++;
        if (c[t] == 2) {
          de[d[t]].erase(t);
        }
        p[i] = t;
        de[d[i]].erase(i);
        d[i]--;
        de[d[i]].insert(i);
        cur--;
        break;
      }
    }
    if (!found) {
      cout << "NO"
           << "\n";
      return;
    }
  }
  cout << "YES"
       << "\n";
  for (int i = 2; i <= n; i++) {
    cout << p[i] << " ";
  }
  cout << "\n";
}
void solve() {
  int q;
  cin >> q;
  while (q--) {
    query();
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout.precision(10);
  solve();
  cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
  return 0;
}

### Prompt

Your challenge is to write 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;
const int N = 1e6 + 1;
const int NN = 1e3 + 1;
const long long MOD = 1e9 + 7;
const long long oo = 1e18 + 7;
const int BASE = 10000;
void query() {
  int n, D;
  cin >> n >> D;
  vector<int> p(n + 1), c(n + 1), d(n + 1);
  vector<set<int>> de(n);
  for (int i = 1; i <= n; i++) {
    p[i] = i - 1;
    c[i] = (i != n ? 1 : 0);
    d[i] = i - 1;
    de[i - 1].insert(i);
  }
  int cur = (n - 1) * n / 2;
  int bot = 0, m = n, s = 1, cnt = 0;
  while (m) {
    bot += min(m, s) * cnt;
    m -= min(m, s);
    s *= 2;
    cnt++;
  }
  if (D > cur || D < bot) {
    cout << "NO"
         << "\n";
    return;
  }
  while (cur > D) {
    int found = 0;
    for (int i = 1; i <= n; i++) {
      if (!c[i] && d[i] >= 2 && de[d[i] - 2].size()) {
        found = 1;
        c[p[i]]--;
        if (c[p[i]] == 1) {
          de[d[p[i]]].insert(p[i]);
        }
        int t = *de[d[i] - 2].begin();
        c[t]++;
        if (c[t] == 2) {
          de[d[t]].erase(t);
        }
        p[i] = t;
        de[d[i]].erase(i);
        d[i]--;
        de[d[i]].insert(i);
        cur--;
        break;
      }
    }
    if (!found) {
      cout << "NO"
           << "\n";
      return;
    }
  }
  cout << "YES"
       << "\n";
  for (int i = 2; i <= n; i++) {
    cout << p[i] << " ";
  }
  cout << "\n";
}
void solve() {
  int q;
  cin >> q;
  while (q--) {
    query();
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout.precision(10);
  solve();
  cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
  return 0;
}

```