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;
class Person {
 public:
  long long int l;
  long long int r, s;
  Person(long long int l, long long int r, long long int s)
      : l(l), r(r), s(s) {}
};
bool operator<(const Person& p1, const Person& p2) {
  if (p1.s != p2.s) return p1.s < p2.s;
  return p1.l > p2.l;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t = 1;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    vector<long long int> v(n, 0);
    priority_queue<Person> p;
    long long int i = 1;
    p.push(Person(0, n - 1, n));
    while (p.size() > 0) {
      Person t = p.top();
      p.pop();
      long long int m;
      long long int r = t.r;
      long long int l = t.l;
      if (r - l + 1 % 2 == 0) {
        m = (r + l - 1) / 2;
      } else {
        m = (r + l) / 2;
      }
      v[m] = i;
      i++;
      if (l <= m - 1) p.push(Person(l, m - 1, m - 1 - l + 1));
      if (r >= m + 1) p.push(Person(m + 1, r, r - m - 1 + 1));
    }
    for (int i = 0; i < n; i++) cout << v[i] << " ";
    cout << endl;
  }
  return 0;
}