#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
const int N = 100005;
void solve() {
  long long n, d;
  cin >> n >> d;
  long long lo = 0, hi = (n * (n - 1)) / 2;
  long long curr = -1;
  for (long long i = 1; i <= n; i++) {
    if (!(i & (i - 1))) curr++;
    lo += curr;
  }
  if (d < lo || d > hi) {
    cout << "NO" << '\n';
    return;
  }
  curr = hi;
  vector<long long> par(n);
  iota((par).begin(), (par).end(), -1);
  vector<long long> dep(n);
  iota((dep).begin(), (dep).end(), 0);
  vector<long long> bad(n);
  vector<long long> cnt(n, 1);
  cnt[n - 1] = 0;
  while (curr > d) {
    long long node = -1;
    for (long long i = 0; i <= n - 1; i++) {
      if (!bad[i] && cnt[i] == 0 && (node == -1 || (dep[i] < dep[node]))) {
        node = i;
      }
    }
    long long p = -1;
    for (long long i = 0; i <= n - 1; i++) {
      if (cnt[i] < 2 && dep[i] == (dep[node] - 2)) {
        p = i;
      }
    }
    if (p == -1) {
      bad[node] = 1;
      continue;
    }
    cnt[par[node]]--;
    par[node] = p;
    cnt[p]++;
    dep[node]--;
    curr--;
  }
  cout << "YES" << '\n';
  for (long long i = 1; i <= n - 1; i++) cout << par[i] + 1 << " ";
  cout << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  cin >> TESTS;
  while (TESTS--) {
    solve();
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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 gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
const int N = 100005;
void solve() {
  long long n, d;
  cin >> n >> d;
  long long lo = 0, hi = (n * (n - 1)) / 2;
  long long curr = -1;
  for (long long i = 1; i <= n; i++) {
    if (!(i & (i - 1))) curr++;
    lo += curr;
  }
  if (d < lo || d > hi) {
    cout << "NO" << '\n';
    return;
  }
  curr = hi;
  vector<long long> par(n);
  iota((par).begin(), (par).end(), -1);
  vector<long long> dep(n);
  iota((dep).begin(), (dep).end(), 0);
  vector<long long> bad(n);
  vector<long long> cnt(n, 1);
  cnt[n - 1] = 0;
  while (curr > d) {
    long long node = -1;
    for (long long i = 0; i <= n - 1; i++) {
      if (!bad[i] && cnt[i] == 0 && (node == -1 || (dep[i] < dep[node]))) {
        node = i;
      }
    }
    long long p = -1;
    for (long long i = 0; i <= n - 1; i++) {
      if (cnt[i] < 2 && dep[i] == (dep[node] - 2)) {
        p = i;
      }
    }
    if (p == -1) {
      bad[node] = 1;
      continue;
    }
    cnt[par[node]]--;
    par[node] = p;
    cnt[p]++;
    dep[node]--;
    curr--;
  }
  cout << "YES" << '\n';
  for (long long i = 1; i <= n - 1; i++) cout << par[i] + 1 << " ";
  cout << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  cin >> TESTS;
  while (TESTS--) {
    solve();
  }
  return 0;
}

```