#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long power(long long a, long long n) {
  a %= 1000000007;
  if (n == 1) return a;
  if (n == 0) return 1;
  if (n % 2)
    return (a * (power((a * a) % 1000000007, n / 2) % 1000000007)) % 1000000007;
  return power((a * a) % 1000000007, n / 2) % 1000000007;
}
const long long inf = (long long)1e18;
long long inverse(long long x) { return power(x, 1000000007 - 2) % 1000000007; }
void solve() {
  long long n, d;
  cin >> n >> d;
  if (d > (n * (n - 1)) / 2) {
    cout << "NO\n";
    return;
  }
  vector<long long> par(n + 5), h(n + 5), cc(n + 5);
  long long cur = (n * (n - 1)) / 2;
  for (long long i = 2; i < n + 1; ++i)
    par[i] = i - 1, h[i] = i - 1, cc[i - 1] = 1;
  set<pair<long long, long long> > pq;
  pq.insert({-n + 1, n});
  while (cur > d) {
    long long x = (*pq.begin()).second, height = -1 * (*pq.begin()).first;
    pq.erase(pq.begin());
    if (height > h[x]) continue;
    long long mxi = -1;
    for (long long i = 1; i < n + 1; ++i) {
      if (cc[i] < 2 && h[i] < h[x] && cur + h[i] + 1 - h[x] >= d) {
        if (mxi == -1 || h[mxi] > h[i]) {
          mxi = i;
        }
      }
    }
    if (mxi == -1) break;
    cur += h[mxi] - h[x] + 1;
    cc[par[x]]--;
    pq.insert({-h[par[x]], par[x]});
    par[x] = mxi;
    cc[mxi]++;
    h[x] = h[mxi] + 1;
  }
  if (cur != d) {
    cout << "NO\n";
    return;
  }
  cout << "YES\n";
  for (long long i = 2; i < n + 1; ++i) cout << par[i] << " ";
  cout << "\n";
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long T = 1;
  cin >> T;
  for (long long t = 1; t < T + 1; ++t) {
    solve();
  }
  return 0;
}

### Prompt

Please formulate 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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long power(long long a, long long n) {
  a %= 1000000007;
  if (n == 1) return a;
  if (n == 0) return 1;
  if (n % 2)
    return (a * (power((a * a) % 1000000007, n / 2) % 1000000007)) % 1000000007;
  return power((a * a) % 1000000007, n / 2) % 1000000007;
}
const long long inf = (long long)1e18;
long long inverse(long long x) { return power(x, 1000000007 - 2) % 1000000007; }
void solve() {
  long long n, d;
  cin >> n >> d;
  if (d > (n * (n - 1)) / 2) {
    cout << "NO\n";
    return;
  }
  vector<long long> par(n + 5), h(n + 5), cc(n + 5);
  long long cur = (n * (n - 1)) / 2;
  for (long long i = 2; i < n + 1; ++i)
    par[i] = i - 1, h[i] = i - 1, cc[i - 1] = 1;
  set<pair<long long, long long> > pq;
  pq.insert({-n + 1, n});
  while (cur > d) {
    long long x = (*pq.begin()).second, height = -1 * (*pq.begin()).first;
    pq.erase(pq.begin());
    if (height > h[x]) continue;
    long long mxi = -1;
    for (long long i = 1; i < n + 1; ++i) {
      if (cc[i] < 2 && h[i] < h[x] && cur + h[i] + 1 - h[x] >= d) {
        if (mxi == -1 || h[mxi] > h[i]) {
          mxi = i;
        }
      }
    }
    if (mxi == -1) break;
    cur += h[mxi] - h[x] + 1;
    cc[par[x]]--;
    pq.insert({-h[par[x]], par[x]});
    par[x] = mxi;
    cc[mxi]++;
    h[x] = h[mxi] + 1;
  }
  if (cur != d) {
    cout << "NO\n";
    return;
  }
  cout << "YES\n";
  for (long long i = 2; i < n + 1; ++i) cout << par[i] << " ";
  cout << "\n";
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long T = 1;
  cin >> T;
  for (long long t = 1; t < T + 1; ++t) {
    solve();
  }
  return 0;
}

```