#include <bits/stdc++.h>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cin.exceptions(cin.failbit);
  int T;
  cin >> T;
  while (T--) {
    long long int t, n, i, j, x, y, m, l, d, k, x1, y1;
    cin >> n;
    vector<pair<long long, long long> > v[n];
    long long int arr[n];
    long long int ans = 1;
    v[n - 1].push_back(make_pair(0, n - 1));
    for (int i = n - 1; i >= 0; i--) {
      sort(v[i].begin(), v[i].end());
      for (int j = 0; j < v[i].size(); j++) {
        x = v[i][j].first;
        y = v[i][j].second;
        d = (x + y) / 2;
        arr[d] = ans++;
        if (x != y) {
          x1 = d - 1;
          if (x1 >= x) v[x1 - x].push_back(make_pair(x, x1));
          y1 = d + 1;
          if (y1 <= y) v[y - y1].push_back(make_pair(y1, y));
        }
      }
    }
    for (int i = 0; i < (n); ++i) cout << arr[i] << ' ';
    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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cin.exceptions(cin.failbit);
  int T;
  cin >> T;
  while (T--) {
    long long int t, n, i, j, x, y, m, l, d, k, x1, y1;
    cin >> n;
    vector<pair<long long, long long> > v[n];
    long long int arr[n];
    long long int ans = 1;
    v[n - 1].push_back(make_pair(0, n - 1));
    for (int i = n - 1; i >= 0; i--) {
      sort(v[i].begin(), v[i].end());
      for (int j = 0; j < v[i].size(); j++) {
        x = v[i][j].first;
        y = v[i][j].second;
        d = (x + y) / 2;
        arr[d] = ans++;
        if (x != y) {
          x1 = d - 1;
          if (x1 >= x) v[x1 - x].push_back(make_pair(x, x1));
          y1 = d + 1;
          if (y1 <= y) v[y - y1].push_back(make_pair(y1, y));
        }
      }
    }
    for (int i = 0; i < (n); ++i) cout << arr[i] << ' ';
    cout << '\n';
  }
}

```