#include <bits/stdc++.h>
const int MAX = 2e5;
using namespace std;
vector<int> vet;
struct comp {
  bool operator()(const pair<int, int>& el1, const pair<int, int>& el2) const {
    if ((el1.second - el1.first + 1) == (el2.second - el2.first + 1)) {
      return el1.first < el2.first;
    }
    return (el1.second - el1.first + 1) >= (el2.second - el2.first + 1);
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vet = vector<int>(n, 0);
    pair<int, int> foo = make_pair(0, n - 1);
    multiset<pair<int, int>, comp> aux;
    multiset<pair<int, int>, comp>::iterator it;
    aux.insert(foo);
    int act = 1;
    while (aux.size() > 0) {
      pair<int, int> here = *(aux.begin());
      it = aux.begin();
      aux.erase(it);
      if (here.second < here.first) continue;
      int mid;
      int r = here.second, l = here.first;
      if ((r - l + 1) % 2 != 0) {
        mid = (l + r) / 2;
      } else {
        mid = (l + r - 1) / 2;
      }
      vet[mid] = act;
      act++;
      pair<int, int> leff = make_pair(l, mid - 1);
      pair<int, int> rightt = make_pair(mid + 1, r);
      aux.insert(leff);
      aux.insert(rightt);
    }
    for (auto x : vet) {
      cout << x << " ";
    }
    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>
const int MAX = 2e5;
using namespace std;
vector<int> vet;
struct comp {
  bool operator()(const pair<int, int>& el1, const pair<int, int>& el2) const {
    if ((el1.second - el1.first + 1) == (el2.second - el2.first + 1)) {
      return el1.first < el2.first;
    }
    return (el1.second - el1.first + 1) >= (el2.second - el2.first + 1);
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vet = vector<int>(n, 0);
    pair<int, int> foo = make_pair(0, n - 1);
    multiset<pair<int, int>, comp> aux;
    multiset<pair<int, int>, comp>::iterator it;
    aux.insert(foo);
    int act = 1;
    while (aux.size() > 0) {
      pair<int, int> here = *(aux.begin());
      it = aux.begin();
      aux.erase(it);
      if (here.second < here.first) continue;
      int mid;
      int r = here.second, l = here.first;
      if ((r - l + 1) % 2 != 0) {
        mid = (l + r) / 2;
      } else {
        mid = (l + r - 1) / 2;
      }
      vet[mid] = act;
      act++;
      pair<int, int> leff = make_pair(l, mid - 1);
      pair<int, int> rightt = make_pair(mid + 1, r);
      aux.insert(leff);
      aux.insert(rightt);
    }
    for (auto x : vet) {
      cout << x << " ";
    }
    cout << '\n';
  }
}

```