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:

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#include <bits/stdc++.h>
using namespace std;
template <class T>
inline bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T &a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
const double EPS = 1e-7;
const int INF = 1 << 30;
const long long int LLINF = 1LL << 60;
const double PI = acos(-1);
const int MOD = 1000000007;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
void dfs(int v, int p, vector<vector<int>> &g, vector<int> &ans) {
  ans[v] = p;
  for (auto u : g[v]) {
    if (u != p) {
      dfs(u, v, g, ans);
    }
  }
}
void solve() {
  int n, d;
  cin >> n >> d;
  if ((n - 1) * n / 2 < d) {
    cout << "NO"
         << "\n";
    return;
  }
  vector<int> v(n, 1);
  int fr = 1, ba = n - 1;
  int sum = (n - 1) * n / 2;
  while (sum > d) {
    if (fr >= ba) {
      cout << "NO"
           << "\n";
      return;
    }
    while (v[ba] == 0) {
      ba--;
    }
    v[ba]--;
    int diff = ba - fr;
    if (sum - d > diff) {
      v[fr]++;
      sum -= diff;
      if (v[fr] == (1 << fr)) {
        fr++;
      }
    } else {
      diff = sum - d;
      v[ba - diff]++;
      break;
    }
  }
  vector<vector<int>> g(n);
  int l = 0, r = 0;
  for (int depth = 0; depth < n - 1; depth++) {
    if (v[depth + 1] == 0) {
      break;
    }
    int cnt = 0;
    for (int now = l; now <= r; now++) {
      g[now].push_back(r + 1 + cnt);
      cnt++;
      if (cnt == v[depth + 1]) {
        break;
      }
      g[now].push_back(r + 1 + cnt);
      cnt++;
      if (cnt == v[depth + 1]) {
        break;
      }
    }
    l = r + 1, r = l + v[depth + 1] - 1;
  }
  vector<int> ans(n);
  dfs(0, -1, g, ans);
  cout << "YES"
       << "\n";
  for (int i = 1; i < n; i++) {
    cout << ans[i] + 1 << ' ';
  }
  cout << "\n";
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}
