#include <bits/stdc++.h>
using namespace std;
struct cmp {
  bool operator()(const pair<long long, pair<long long, long long>> &p1,
                  const pair<long long, pair<long long, long long>> &p2) {
    if (p1.first == p2.first) {
      return (p1.second.first > p2.second.first);
    }
    return p1.first < p2.first;
    return true;
  }
};
signed main() {
  std ::ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    set<pair<long long, pair<long long, long long>>, cmp> segs;
    segs.insert({n + 1, {1, n}});
    vector<long long> answer(n + 1);
    long long ind = 0;
    while (ind != n) {
      auto p = segs.rbegin();
      auto pp = *p;
      long long len = pp.first;
      long long l = pp.second.first, r = pp.second.second;
      long long mid = (l + r) / 2;
      answer[mid] = ++ind;
      segs.erase(prev(segs.end()));
      long long len1 = mid - l;
      long long len2 = r - (mid + 1) + 1;
      segs.insert({len1, {l, mid - 1}});
      segs.insert({len2, {mid + 1, r}});
    }
    for (long long i = 1; i <= n; i++) cout << answer[i] << ' ';
    cout << '\n';
  }
}

### Prompt

Develop a solution in CPP to the problem described below:

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 cmp {
  bool operator()(const pair<long long, pair<long long, long long>> &p1,
                  const pair<long long, pair<long long, long long>> &p2) {
    if (p1.first == p2.first) {
      return (p1.second.first > p2.second.first);
    }
    return p1.first < p2.first;
    return true;
  }
};
signed main() {
  std ::ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    set<pair<long long, pair<long long, long long>>, cmp> segs;
    segs.insert({n + 1, {1, n}});
    vector<long long> answer(n + 1);
    long long ind = 0;
    while (ind != n) {
      auto p = segs.rbegin();
      auto pp = *p;
      long long len = pp.first;
      long long l = pp.second.first, r = pp.second.second;
      long long mid = (l + r) / 2;
      answer[mid] = ++ind;
      segs.erase(prev(segs.end()));
      long long len1 = mid - l;
      long long len2 = r - (mid + 1) + 1;
      segs.insert({len1, {l, mid - 1}});
      segs.insert({len2, {mid + 1, r}});
    }
    for (long long i = 1; i <= n; i++) cout << answer[i] << ' ';
    cout << '\n';
  }
}

```