#include <bits/stdc++.h>
using namespace std;
const long long MxN = 5e3 + 5;
long long n, d, nc[MxN], p[MxN];
vector<long long> w[MxN];
void solve() {
  cin >> n >> d;
  string yes = "YES\n", no = "NO\n";
  long long i, ub_d, lb_d, temp;
  ub_d = (n * (n - 1)) / 2;
  lb_d = 0;
  for (i = 1; i <= n; i++) {
    temp = i;
    while (temp > 0) {
      temp /= 2;
      if (temp != 0) lb_d++;
    }
  }
  if (!(lb_d <= d && d <= ub_d)) {
    cout << no;
    return;
  }
  for (i = 0; i < n; i++) {
    nc[i] = 1;
    p[i] = i - 1;
    w[i] = {i};
  }
  long long s = ub_d, u = n - 1, a, v;
  while (s > d) {
    nc[u] = 0;
    a = 0;
    while (a + 1 < u && !w[a].size()) {
      a++;
    }
    if (u - a - 1 > s - d) {
      p[u] = u - 1 - (s - d);
      break;
    }
    v = w[a].back();
    p[u] = v;
    nc[v]++;
    s -= (u - a - 1);
    if (nc[v] >= 2) {
      w[a].pop_back();
    }
    w[a + 1].push_back(u);
    u--;
  }
  cout << yes;
  for (i = 1; i < n; i++) {
    cout << p[i] + 1 << " ";
  }
  cout << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t;
  cin >> t;
  while (t--) 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;
const long long MxN = 5e3 + 5;
long long n, d, nc[MxN], p[MxN];
vector<long long> w[MxN];
void solve() {
  cin >> n >> d;
  string yes = "YES\n", no = "NO\n";
  long long i, ub_d, lb_d, temp;
  ub_d = (n * (n - 1)) / 2;
  lb_d = 0;
  for (i = 1; i <= n; i++) {
    temp = i;
    while (temp > 0) {
      temp /= 2;
      if (temp != 0) lb_d++;
    }
  }
  if (!(lb_d <= d && d <= ub_d)) {
    cout << no;
    return;
  }
  for (i = 0; i < n; i++) {
    nc[i] = 1;
    p[i] = i - 1;
    w[i] = {i};
  }
  long long s = ub_d, u = n - 1, a, v;
  while (s > d) {
    nc[u] = 0;
    a = 0;
    while (a + 1 < u && !w[a].size()) {
      a++;
    }
    if (u - a - 1 > s - d) {
      p[u] = u - 1 - (s - d);
      break;
    }
    v = w[a].back();
    p[u] = v;
    nc[v]++;
    s -= (u - a - 1);
    if (nc[v] >= 2) {
      w[a].pop_back();
    }
    w[a + 1].push_back(u);
    u--;
  }
  cout << yes;
  for (i = 1; i < n; i++) {
    cout << p[i] + 1 << " ";
  }
  cout << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```