#include <bits/stdc++.h>
using namespace std;
struct CustomCompare {
  bool operator()(const pair<int, int>& b, const pair<int, int>& a) {
    if (abs(a.second - a.first) > abs(b.second - b.first))
      return true;
    else if (abs(a.second - a.first) == abs(b.second - b.first))
      if (a.first < b.first) return true;
    return false;
  }
};
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    pair<int, int> p = make_pair(0, n - 1);
    priority_queue<pair<int, int>, vector<pair<int, int> >, CustomCompare> q;
    vector<int> ans(n, 0);
    q.push(p);
    int i = 0;
    while (!q.empty()) {
      i += 1;
      pair<int, int> temp = q.top();
      q.pop();
      if (temp.first > temp.second) continue;
      pair<int, int> p1, p2;
      int mid = (temp.first + temp.second) / 2;
      if (temp.first == temp.second) {
        ans[temp.first] = i;
        continue;
      } else {
        ans[mid] = i;
      }
      if (temp.first < mid) {
        q.push(make_pair(temp.first, mid - 1));
      }
      if (temp.second > mid) {
        q.push(make_pair(mid + 1, temp.second));
      }
    }
    for (int j = 0; j < n; j++) cout << ans[j] << " ";
    cout << "\n";
  }
}

### Prompt

Your task is to create a cpp solution to 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 CustomCompare {
  bool operator()(const pair<int, int>& b, const pair<int, int>& a) {
    if (abs(a.second - a.first) > abs(b.second - b.first))
      return true;
    else if (abs(a.second - a.first) == abs(b.second - b.first))
      if (a.first < b.first) return true;
    return false;
  }
};
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    pair<int, int> p = make_pair(0, n - 1);
    priority_queue<pair<int, int>, vector<pair<int, int> >, CustomCompare> q;
    vector<int> ans(n, 0);
    q.push(p);
    int i = 0;
    while (!q.empty()) {
      i += 1;
      pair<int, int> temp = q.top();
      q.pop();
      if (temp.first > temp.second) continue;
      pair<int, int> p1, p2;
      int mid = (temp.first + temp.second) / 2;
      if (temp.first == temp.second) {
        ans[temp.first] = i;
        continue;
      } else {
        ans[mid] = i;
      }
      if (temp.first < mid) {
        q.push(make_pair(temp.first, mid - 1));
      }
      if (temp.second > mid) {
        q.push(make_pair(mid + 1, temp.second));
      }
    }
    for (int j = 0; j < n; j++) cout << ans[j] << " ";
    cout << "\n";
  }
}

```