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 SubArr {
 public:
  int start_idx;
  int end_idx;
  int len;
  SubArr(int st, int ed, int l) : start_idx(st), end_idx(ed), len(l){};
};
bool operator<(const SubArr& l, const SubArr& r) {
  if (l.len < r.len)
    return true;
  else if (l.len == r.len)
    return (l.start_idx > r.start_idx);
  return false;
}
void solve(int n) {
  int* arr = new int[n + 1];
  memset(arr, 0, sizeof(int) * (n + 1));
  priority_queue<SubArr> q;
  q.emplace(1, n, n);
  for (int i = 1; i <= n; i++) {
    SubArr here = q.top();
    q.pop();
    int mid = (here.start_idx + here.end_idx) / 2;
    arr[mid] = i;
    int len = (mid - here.start_idx);
    if (len >= 1) q.emplace(here.start_idx, mid - 1, len);
    len = (here.end_idx - mid);
    if (len >= 1) q.emplace(mid + 1, here.end_idx, len);
  }
  for (int i = 1; i <= n; i++) cout << arr[i] << " ";
  cout << endl;
  delete arr;
}
int main() {
  int t;
  cin >> t;
  int n;
  for (int i = 0; i < t; i++) {
    cin >> n;
    solve(n);
  }
  return 0;
}