#include <bits/stdc++.h>
using namespace std;
struct group {
  group(int s, int e, int g) : start(s), end(e), gap(g) {}
  int start;
  int end;
  int gap;
};
struct cmp {
  bool operator()(group a, group b) {
    if (a.gap == b.gap) return a.start > b.start;
    return a.gap < b.gap;
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int T;
  cin >> T;
  int n;
  priority_queue<group, vector<group>, cmp> pq;
  while (T--) {
    cin >> n;
    group x(1, n, n - 1);
    group y(1, n, n - 1);
    vector<int> v(n + 1);
    pq.push(x);
    int mid;
    for (int i = 1; i <= n; i++) {
      y = pq.top();
      pq.pop();
      if (y.gap == 0) {
        v[y.start] = i;
        continue;
      }
      if (y.gap == 1) {
        v[y.start] = i;
        x.start = y.end;
        x.end = y.end;
        x.gap = 0;
        pq.push(x);
        continue;
      }
      if (i % 2 == 0) {
        v[(y.start + y.end) / 2] = i;
        mid = (y.start + y.end) / 2;
      } else {
        v[(y.start + y.end) / 2] = i;
        mid = (y.start + y.end) / 2;
      }
      x.start = y.start;
      x.end = mid - 1;
      x.gap = x.end - x.start;
      pq.push(x);
      x.start = mid + 1;
      x.end = y.end;
      x.gap = x.end - x.start;
      pq.push(x);
    }
    for (int i = 1; i <= n; i++) cout << v[i] << " ";
    cout << "\n";
  }
  return 0;
}

### 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 group {
  group(int s, int e, int g) : start(s), end(e), gap(g) {}
  int start;
  int end;
  int gap;
};
struct cmp {
  bool operator()(group a, group b) {
    if (a.gap == b.gap) return a.start > b.start;
    return a.gap < b.gap;
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int T;
  cin >> T;
  int n;
  priority_queue<group, vector<group>, cmp> pq;
  while (T--) {
    cin >> n;
    group x(1, n, n - 1);
    group y(1, n, n - 1);
    vector<int> v(n + 1);
    pq.push(x);
    int mid;
    for (int i = 1; i <= n; i++) {
      y = pq.top();
      pq.pop();
      if (y.gap == 0) {
        v[y.start] = i;
        continue;
      }
      if (y.gap == 1) {
        v[y.start] = i;
        x.start = y.end;
        x.end = y.end;
        x.gap = 0;
        pq.push(x);
        continue;
      }
      if (i % 2 == 0) {
        v[(y.start + y.end) / 2] = i;
        mid = (y.start + y.end) / 2;
      } else {
        v[(y.start + y.end) / 2] = i;
        mid = (y.start + y.end) / 2;
      }
      x.start = y.start;
      x.end = mid - 1;
      x.gap = x.end - x.start;
      pq.push(x);
      x.start = mid + 1;
      x.end = y.end;
      x.gap = x.end - x.start;
      pq.push(x);
    }
    for (int i = 1; i <= n; i++) cout << v[i] << " ";
    cout << "\n";
  }
  return 0;
}

```