#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
long long T, n, a[N];
vector<pair<int, int> > v[N];
int main() {
  cin >> T;
  while (T--) {
    cin >> n;
    v[n].push_back({1, n});
    int cur = 1;
    for (int i = n; i > 0; i--) {
      sort(v[i].begin(), v[i].end());
      for (int j = 0; j < v[i].size(); j++) {
        int l = v[i][j].first;
        int r = v[i][j].second;
        int pos = (r - l + 1) % 2 == 0 ? (r + l - 1) / 2 : (l + r) / 2;
        a[pos] = cur++;
        if (l == r) {
        } else if (l + 1 == r) {
          v[1].push_back({r, r});
        } else {
          if ((r - l + 1) % 2 == 0) {
            v[r - pos].push_back({pos + 1, r});
            v[pos - l].push_back({l, pos - 1});
          } else {
            v[pos - l].push_back({l, pos - 1});
            v[r - pos].push_back({pos + 1, r});
          }
        }
      }
      v[i].clear();
    }
    for (int i = 1; i <= n; i++) cout << a[i] << " ";
    cout << endl;
  }
}

### Prompt

In Cpp, your task is to solve 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 = 200005;
long long T, n, a[N];
vector<pair<int, int> > v[N];
int main() {
  cin >> T;
  while (T--) {
    cin >> n;
    v[n].push_back({1, n});
    int cur = 1;
    for (int i = n; i > 0; i--) {
      sort(v[i].begin(), v[i].end());
      for (int j = 0; j < v[i].size(); j++) {
        int l = v[i][j].first;
        int r = v[i][j].second;
        int pos = (r - l + 1) % 2 == 0 ? (r + l - 1) / 2 : (l + r) / 2;
        a[pos] = cur++;
        if (l == r) {
        } else if (l + 1 == r) {
          v[1].push_back({r, r});
        } else {
          if ((r - l + 1) % 2 == 0) {
            v[r - pos].push_back({pos + 1, r});
            v[pos - l].push_back({l, pos - 1});
          } else {
            v[pos - l].push_back({l, pos - 1});
            v[r - pos].push_back({pos + 1, r});
          }
        }
      }
      v[i].clear();
    }
    for (int i = 1; i <= n; i++) cout << a[i] << " ";
    cout << endl;
  }
}

```