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>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long parent[200005];
long long dis[200005];
long long jura[200005];
long long isbad[200005];
long long istake[200005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long i, j, k, l, n, m;
  long long t;
  cin >> t;
  while (t--) {
    cin >> n >> m;
    memset(isbad, 0, sizeof(long long) * (n + 1));
    memset(dis, 0, sizeof(long long) * (n + 1));
    memset(jura, 0, sizeof(long long) * (n + 1));
    memset(istake, 0, sizeof(long long) * (n + 1));
    long long pre = n - 1;
    long long maxi = (pre * (pre + 1)) / 2;
    long long mini = 0;
    long long cnt = 1;
    while (pre > 0) {
      if (pre >= (1LL << cnt)) {
        mini += ((1LL << cnt) * cnt);
        pre -= (1LL << cnt);
        cnt++;
      } else {
        mini += (pre * cnt);
        pre = 0;
        cnt++;
      }
    }
    if (m > maxi || m < mini) {
      cout << "NO\n";
    } else {
      deque<long long> leafs;
      jura[1] = 1;
      for (i = 2; i < n; i++) {
        parent[i] = i - 1;
        jura[i] = 1;
        dis[i] = i - 1;
      }
      dis[n] = n - 1;
      parent[n] = n - 1;
      long long sum = ((n - 1) * n) / 2;
      leafs.push_back(n);
      while (sum > m && !leafs.empty()) {
        long long kola = leafs[0];
        if (istake[kola] == 1) {
          leafs.pop_front();
          continue;
        }
        long long needed = dis[kola] - 1;
        int flag = 0;
        for (i = 1; i <= n; i++) {
          if (i != kola && (dis[i] == (needed - 1)) && isbad[i] == 0 &&
              jura[i] < 2) {
            jura[parent[kola]]--;
            isbad[parent[kola]] = 0;
            if (jura[parent[kola]] == 0) {
              leafs.push_back(parent[kola]);
            }
            parent[kola] = i;
            jura[i]++;
            if (jura[i] == 2) isbad[i] = 1;
            dis[kola] = needed;
            sum--;
            flag = 1;
            break;
          }
        }
        if (flag == 0) istake[kola] = 1;
      }
      if (sum != m) {
        cout << "NO\n";
      } else {
        cout << "YES\n";
        for (i = 2; i <= n; i++) cout << parent[i] << " ";
        cout << "\n";
      }
    }
  }
}