#include <bits/stdc++.h>
using namespace std;
int ans[200005];
int cont;
void solve() {
  int n;
  cin >> n;
  cont = 0;
  memset(ans, 0, sizeof ans);
  priority_queue<pair<int, int>> p;
  p.push({n, -1});
  while (p.size()) {
    if (cont > n) break;
    pair<int, int> now = p.top();
    p.pop();
    int len = now.first;
    int start = now.second * -1;
    int endd = start + len - 1;
    cont++;
    if (len & 1) {
      ans[(start + endd) / 2] = cont;
      int newlen = (len - 1) / 2;
      if (newlen == 0) continue;
      int firstStart = start * -1;
      int secondStart = (start + endd) / 2 + 1;
      secondStart *= -1;
      p.push({newlen, firstStart});
      p.push({newlen, secondStart});
    } else {
      ans[(start + endd - 1) / 2] = cont;
      int firstlen = (len - 1) / 2;
      int secondlen = len / 2;
      int firstStart = start * -1;
      int secondStart = (start + endd - 1) / 2 + 1;
      secondStart *= -1;
      if (firstlen != 0) p.push({firstlen, firstStart});
      if (secondlen != 0) p.push({secondlen, secondStart});
    }
  }
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### 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;
int ans[200005];
int cont;
void solve() {
  int n;
  cin >> n;
  cont = 0;
  memset(ans, 0, sizeof ans);
  priority_queue<pair<int, int>> p;
  p.push({n, -1});
  while (p.size()) {
    if (cont > n) break;
    pair<int, int> now = p.top();
    p.pop();
    int len = now.first;
    int start = now.second * -1;
    int endd = start + len - 1;
    cont++;
    if (len & 1) {
      ans[(start + endd) / 2] = cont;
      int newlen = (len - 1) / 2;
      if (newlen == 0) continue;
      int firstStart = start * -1;
      int secondStart = (start + endd) / 2 + 1;
      secondStart *= -1;
      p.push({newlen, firstStart});
      p.push({newlen, secondStart});
    } else {
      ans[(start + endd - 1) / 2] = cont;
      int firstlen = (len - 1) / 2;
      int secondlen = len / 2;
      int firstStart = start * -1;
      int secondStart = (start + endd - 1) / 2 + 1;
      secondStart *= -1;
      if (firstlen != 0) p.push({firstlen, firstStart});
      if (secondlen != 0) p.push({secondlen, secondStart});
    }
  }
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```