#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
inline int gcd(int a, int b) {
  while (a != 0 && b != 0) {
    if (a > b) {
      a %= b;
    } else {
      b %= a;
    }
  }
  return max(a, b);
}
struct compare {
  bool operator()(pair<int, int> const& p1, pair<int, int> const& p2) {
    int size = p1.second - p1.first + 1;
    int size2 = p2.second - p2.first + 1;
    if (size2 > size) {
      return 1;
    } else if (size2 == size) {
      if (p1.first > p2.first) {
        return 1;
      }
    }
    return 0;
  }
};
void solve() {
  int n;
  cin >> n;
  vector<int> ans(n);
  int k = 1;
  priority_queue<pair<int, int>, vector<pair<int, int> >, compare> pq;
  pq.push({0, n - 1});
  while (!pq.empty()) {
    pair<int, int> temp = pq.top();
    pq.pop();
    int l = temp.first, r = temp.second;
    int size = (l - r + 1);
    int mid = l + (r - l) / 2;
    ans[mid] = k;
    k++;
    if (mid - 1 >= l) pq.push({l, mid - 1});
    if (r >= mid + 1) pq.push({mid + 1, r});
  }
  for (auto it : ans) cout << it << " ";
  cout << "\n";
  ;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Generate 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;
const int N = 100005;
inline int gcd(int a, int b) {
  while (a != 0 && b != 0) {
    if (a > b) {
      a %= b;
    } else {
      b %= a;
    }
  }
  return max(a, b);
}
struct compare {
  bool operator()(pair<int, int> const& p1, pair<int, int> const& p2) {
    int size = p1.second - p1.first + 1;
    int size2 = p2.second - p2.first + 1;
    if (size2 > size) {
      return 1;
    } else if (size2 == size) {
      if (p1.first > p2.first) {
        return 1;
      }
    }
    return 0;
  }
};
void solve() {
  int n;
  cin >> n;
  vector<int> ans(n);
  int k = 1;
  priority_queue<pair<int, int>, vector<pair<int, int> >, compare> pq;
  pq.push({0, n - 1});
  while (!pq.empty()) {
    pair<int, int> temp = pq.top();
    pq.pop();
    int l = temp.first, r = temp.second;
    int size = (l - r + 1);
    int mid = l + (r - l) / 2;
    ans[mid] = k;
    k++;
    if (mid - 1 >= l) pq.push({l, mid - 1});
    if (r >= mid + 1) pq.push({mid + 1, r});
  }
  for (auto it : ans) cout << it << " ";
  cout << "\n";
  ;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

```