#include <bits/stdc++.h>
using namespace std;
struct myComparator {
  bool operator()(pair<int, pair<int, int> > a, pair<int, pair<int, int> > b) {
    if (a.first == b.first) {
      return a.second.first > b.second.first;
    }
    return a.first < b.first;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int tc;
  cin >> tc;
  while (tc--) {
    int n;
    cin >> n;
    vector<int> ans(n + 1);
    priority_queue<pair<int, pair<int, int> >,
                   vector<pair<int, pair<int, int> > >, myComparator>
        max_heap;
    max_heap.push({n, {1, n}});
    int pos = 1;
    while (!max_heap.empty()) {
      pair<int, pair<int, int> > val = max_heap.top();
      max_heap.pop();
      int l = val.second.first;
      int r = val.second.second;
      int mid = (l + r) / 2;
      ans[mid] = pos;
      if (l <= mid - 1) max_heap.push({mid - l, {l, mid - 1}});
      if (mid + 1 <= r) max_heap.push({r - mid, {mid + 1, r}});
      ++pos;
    }
    for (int i = 1; i <= n; ++i) {
      cout << ans[i] << " ";
    }
    cout << endl;
  }
  cout << endl;
}

### Prompt

Create a solution in CPP for the following 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 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
struct myComparator {
  bool operator()(pair<int, pair<int, int> > a, pair<int, pair<int, int> > b) {
    if (a.first == b.first) {
      return a.second.first > b.second.first;
    }
    return a.first < b.first;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int tc;
  cin >> tc;
  while (tc--) {
    int n;
    cin >> n;
    vector<int> ans(n + 1);
    priority_queue<pair<int, pair<int, int> >,
                   vector<pair<int, pair<int, int> > >, myComparator>
        max_heap;
    max_heap.push({n, {1, n}});
    int pos = 1;
    while (!max_heap.empty()) {
      pair<int, pair<int, int> > val = max_heap.top();
      max_heap.pop();
      int l = val.second.first;
      int r = val.second.second;
      int mid = (l + r) / 2;
      ans[mid] = pos;
      if (l <= mid - 1) max_heap.push({mid - l, {l, mid - 1}});
      if (mid + 1 <= r) max_heap.push({r - mid, {mid + 1, r}});
      ++pos;
    }
    for (int i = 1; i <= n; ++i) {
      cout << ans[i] << " ";
    }
    cout << endl;
  }
  cout << endl;
}

```