#include <bits/stdc++.h>
using namespace std;
const int N = 5004;
unordered_map<int, int> g[N];
int par[N];
int dep[N];
vector<int> num[N];
void solve() {
  int n, d;
  cin >> n >> d;
  int mn = 0;
  int mx = (n * (n - 1)) / 2;
  int lev = 0;
  int nodes = n;
  int sub = 1;
  while (nodes > 0) {
    mn += (lev * min(sub, nodes));
    nodes -= min(nodes, sub);
    sub <<= 1;
    lev++;
  }
  if (d < mn || d > mx) {
    cout << "NO\n";
    return;
  }
  for (int i = 1; i <= n; i++) g[i].clear();
  for (int i = 1; i <= n; i++) num[i].clear();
  dep[1] = 0;
  num[0].push_back(1);
  par[1] = -1;
  int run = 0;
  for (int i = 2; i <= n; i++) {
    g[i][i / 2] = 1;
    g[i / 2][i] = 1;
    par[i] = i / 2;
    dep[i] = dep[i / 2] + 1;
    run += dep[i];
    num[dep[i]].push_back(i);
  }
  while (run < d) {
    for (int i = n; i >= 0; i--) {
      if (num[i].size() > 1) {
        auto x = num[i].back();
        num[i].pop_back();
        num[i + 1].push_back(x);
        int pp = par[x];
        g[pp].erase(x);
        g[x].erase(pp);
        int np = -1;
        for (auto z : num[i]) {
          if (g[z].size() <= 2) {
            np = z;
            break;
          }
        }
        g[np][x] = 1;
        g[x][np] = 1;
        run++;
        par[x] = np;
        break;
      }
    }
  }
  cout << "YES\n";
  for (int i = 2; i <= n; i++) cout << par[i] << " ";
  cout << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) solve();
}

### 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 = 5004;
unordered_map<int, int> g[N];
int par[N];
int dep[N];
vector<int> num[N];
void solve() {
  int n, d;
  cin >> n >> d;
  int mn = 0;
  int mx = (n * (n - 1)) / 2;
  int lev = 0;
  int nodes = n;
  int sub = 1;
  while (nodes > 0) {
    mn += (lev * min(sub, nodes));
    nodes -= min(nodes, sub);
    sub <<= 1;
    lev++;
  }
  if (d < mn || d > mx) {
    cout << "NO\n";
    return;
  }
  for (int i = 1; i <= n; i++) g[i].clear();
  for (int i = 1; i <= n; i++) num[i].clear();
  dep[1] = 0;
  num[0].push_back(1);
  par[1] = -1;
  int run = 0;
  for (int i = 2; i <= n; i++) {
    g[i][i / 2] = 1;
    g[i / 2][i] = 1;
    par[i] = i / 2;
    dep[i] = dep[i / 2] + 1;
    run += dep[i];
    num[dep[i]].push_back(i);
  }
  while (run < d) {
    for (int i = n; i >= 0; i--) {
      if (num[i].size() > 1) {
        auto x = num[i].back();
        num[i].pop_back();
        num[i + 1].push_back(x);
        int pp = par[x];
        g[pp].erase(x);
        g[x].erase(pp);
        int np = -1;
        for (auto z : num[i]) {
          if (g[z].size() <= 2) {
            np = z;
            break;
          }
        }
        g[np][x] = 1;
        g[x][np] = 1;
        run++;
        par[x] = np;
        break;
      }
    }
  }
  cout << "YES\n";
  for (int i = 2; i <= n; i++) cout << par[i] << " ";
  cout << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) solve();
}

```