Problem: You are given an array a of length n consisting of zeros. You perform n actions with this array: during the i-th action, the following sequence of operations appears:

  1. Choose the maximum by length subarray (continuous subsegment) consisting only of zeros, among all such segments choose the leftmost one; 
  2. Let this segment be [l; r]. If r-l+1 is odd (not divisible by 2) then assign (set) a[(l+r)/(2)] := i (where i is the number of the current action), otherwise (if r-l+1 is even) assign (set) a[(l+r-1)/(2)] := i. 



Consider the array a of length 5 (initially a=[0, 0, 0, 0, 0]). Then it changes as follows:

  1. Firstly, we choose the segment [1; 5] and assign a[3] := 1, so a becomes [0, 0, 1, 0, 0]; 
  2. then we choose the segment [1; 2] and assign a[1] := 2, so a becomes [2, 0, 1, 0, 0]; 
  3. then we choose the segment [4; 5] and assign a[4] := 3, so a becomes [2, 0, 1, 3, 0]; 
  4. then we choose the segment [2; 2] and assign a[2] := 4, so a becomes [2, 4, 1, 3, 0]; 
  5. and at last we choose the segment [5; 5] and assign a[5] := 5, so a becomes [2, 4, 1, 3, 5]. 



Your task is to find the array a of length n after performing all n actions. Note that the answer exists and unique.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.

The only line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a.

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).

Output

For each test case, print the answer — the array a of length n after performing n actions described in the problem statement. Note that the answer exists and unique.

Example

Input


6
1
2
3
4
5
6


Output


1 
1 2 
2 1 3 
3 1 2 4 
2 4 1 3 5 
3 4 1 5 2 6 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int md = 1e9 + 7;
const int MX = 2e5 + 5;
const int INF = 1e18;
const long double PI = 4 * atan((long double)1);
int power_md(int a, int n) {
  int res = 1;
  while (n) {
    if (n % 2) res = (res % md * a % md) % md;
    a = (a % md * a % md) % md;
    n /= 2;
  }
  res %= md;
  return res;
}
int power(int a, int n) {
  int res = 1;
  while (n) {
    if (n % 2) res *= a;
    a = a * a;
    n /= 2;
  }
  return res;
}
int abst(int a) { return ((a < 0) ? (-1 * a) : (a)); }
class cmp_set {
 public:
  bool operator()(pair<int, int> a, pair<int, int> b) {
    if (a.first != b.first) return (a.first > b.first);
    return (a.second < b.second);
  }
};
vector<string> vec_splitter(string s) {
  s += ',';
  vector<string> res;
  while (!s.empty()) {
    res.push_back(s.substr(0, s.find(',')));
    s = s.substr(s.find(',') + 1);
  }
  return res;
}
void debug_out(vector<string> __attribute__((unused)) args,
               __attribute__((unused)) int idx,
               __attribute__((unused)) int LINE_NUM) {
  cerr << endl;
}
template <typename Head, typename... Tail>
void debug_out(vector<string> args, int idx, int LINE_NUM, Head H, Tail... T) {
  if (idx > 0)
    cerr << ", ";
  else
    cerr << "Line(" << LINE_NUM << ") ";
  stringstream ss;
  ss << H;
  cerr << args[idx] << " = " << ss.str();
  debug_out(args, idx + 1, LINE_NUM, T...);
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    set<pair<int, int>, cmp_set> st;
    st.insert({n, 1});
    int res[n + 1];
    int cnt = 1;
    while (!st.empty()) {
      pair<int, int> p = *(st.begin());
      st.erase(p);
      int l = p.second;
      int r = p.second + p.first - 1;
      int in = (l + r) / 2;
      res[in] = cnt;
      cnt++;
      if ((in - 1) >= l) st.insert({in - l, l});
      if ((in + 1) <= r) st.insert({r - in, in + 1});
    }
    for (int i = 1; i <= n; i++) cout << res[i] << " ";
    cout << "\n";
  }
  return 0;
}