#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 0;
  cin >> t;
  while (t--) {
    long long int n = 0, m = 0, k = 0, i = 0, j = 0, p = 0, q = 0, x = 0, y = 0,
                  z = 0, cnt = 0;
    string s;
    bool flag = false;
    cin >> n;
    long long int ans[n + 1];
    for (i = 1; i <= n; i++) ans[i] = -1;
    priority_queue<pair<long long int, pair<long long int, long long int>>> qq;
    qq.push(make_pair(n, make_pair(1, n)));
    i = 1;
    long long int sz;
    while (!qq.empty()) {
      sz = qq.top().first;
      x = abs(qq.top().second.first);
      y = abs(qq.top().second.second);
      qq.pop();
      z = (x + y) / 2;
      if (ans[z] == -1) ans[z] = i, i++;
      if (y > x) {
        if ((y - x + 1) % 2 == 0) {
          qq.push(make_pair(y - z, make_pair(-(z + 1), max(y, z + 1))));
          qq.push(make_pair(z - x, make_pair(-x, max(z - 1, x))));
        } else {
          qq.push(make_pair(z - x, make_pair(-x, max(z - 1, x))));
          qq.push(make_pair(y - z, make_pair(-(z + 1), max(y, z + 1))));
        }
      }
    }
    for (i = 1; i <= n; i++) cout << ans[i] << " ";
    cout << endl;
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 0;
  cin >> t;
  while (t--) {
    long long int n = 0, m = 0, k = 0, i = 0, j = 0, p = 0, q = 0, x = 0, y = 0,
                  z = 0, cnt = 0;
    string s;
    bool flag = false;
    cin >> n;
    long long int ans[n + 1];
    for (i = 1; i <= n; i++) ans[i] = -1;
    priority_queue<pair<long long int, pair<long long int, long long int>>> qq;
    qq.push(make_pair(n, make_pair(1, n)));
    i = 1;
    long long int sz;
    while (!qq.empty()) {
      sz = qq.top().first;
      x = abs(qq.top().second.first);
      y = abs(qq.top().second.second);
      qq.pop();
      z = (x + y) / 2;
      if (ans[z] == -1) ans[z] = i, i++;
      if (y > x) {
        if ((y - x + 1) % 2 == 0) {
          qq.push(make_pair(y - z, make_pair(-(z + 1), max(y, z + 1))));
          qq.push(make_pair(z - x, make_pair(-x, max(z - 1, x))));
        } else {
          qq.push(make_pair(z - x, make_pair(-x, max(z - 1, x))));
          qq.push(make_pair(y - z, make_pair(-(z + 1), max(y, z + 1))));
        }
      }
    }
    for (i = 1; i <= n; i++) cout << ans[i] << " ";
    cout << endl;
  }
  return 0;
}

```