#include <bits/stdc++.h>
using namespace std;
long long int powe(long long int a, long long int b) {
  long long int ans = 1;
  for (; b; b >>= 1) {
    if (b & 1) ans = ans * a;
    a = a * a;
  }
  return ans;
}
class Comp {
 public:
  bool operator()(pair<long long int, long long int> a,
                  pair<long long int, long long int> b) {
    if (a.first < b.first || (a.first == b.first && a.second > b.second)) {
      return true;
    }
    return false;
  }
};
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> a(n, 0);
  priority_queue<pair<long long int, long long int>,
                 vector<pair<long long int, long long int>>, Comp>
      vp;
  vp.push({n, 0});
  long long int ind = 0, indd = n - 1;
  long long int mid, cur = 1;
  while (!vp.empty()) {
    pair<long long int, long long int> p = vp.top();
    vp.pop();
    ind = p.second;
    indd = ind + p.first - 1;
    long long int len = indd - ind + 1;
    long long int pp = indd - ind + 1;
    if (pp & 1)
      mid = (ind + indd) / 2;
    else {
      mid = (ind + indd - 1) / 2;
    }
    a[mid] = cur++;
    if ((indd - mid) > 0) vp.push({(indd - mid), mid + 1});
    if (mid - ind > 0) {
      vp.push({(mid - ind), ind});
    }
  }
  for (long long int i = 0; i < n; i++) cout << a[i] << " ";
  cout << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
long long int powe(long long int a, long long int b) {
  long long int ans = 1;
  for (; b; b >>= 1) {
    if (b & 1) ans = ans * a;
    a = a * a;
  }
  return ans;
}
class Comp {
 public:
  bool operator()(pair<long long int, long long int> a,
                  pair<long long int, long long int> b) {
    if (a.first < b.first || (a.first == b.first && a.second > b.second)) {
      return true;
    }
    return false;
  }
};
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> a(n, 0);
  priority_queue<pair<long long int, long long int>,
                 vector<pair<long long int, long long int>>, Comp>
      vp;
  vp.push({n, 0});
  long long int ind = 0, indd = n - 1;
  long long int mid, cur = 1;
  while (!vp.empty()) {
    pair<long long int, long long int> p = vp.top();
    vp.pop();
    ind = p.second;
    indd = ind + p.first - 1;
    long long int len = indd - ind + 1;
    long long int pp = indd - ind + 1;
    if (pp & 1)
      mid = (ind + indd) / 2;
    else {
      mid = (ind + indd - 1) / 2;
    }
    a[mid] = cur++;
    if ((indd - mid) > 0) vp.push({(indd - mid), mid + 1});
    if (mid - ind > 0) {
      vp.push({(mid - ind), ind});
    }
  }
  for (long long int i = 0; i < n; i++) cout << a[i] << " ";
  cout << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

```