#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
struct cmp {
  bool operator()(const pair<long long, pair<long long, long long>> &a,
                  const pair<long long, pair<long long, long long>> &b) const {
    if (a.first != b.first)
      return a.first > b.first;
    else {
      return a.second.first < b.second.first;
    }
  }
};
void solver(long long n) {
  vector<long long> ans(n + 1, 0);
  set<pair<long long, pair<long long, long long>>, cmp> d;
  d.insert({n, {0, n - 1}});
  long long ctr = 1;
  while (d.size() != 0) {
    pair<long long, pair<long long, long long>> pp = *d.begin();
    d.erase(d.begin());
    if (pp.second.first < 0 || pp.second.second < 0 ||
        pp.second.first > pp.second.second)
      continue;
    long long mid = (pp.second.first + pp.second.second) / 2;
    ans[mid] = ctr++;
    d.insert(make_pair((mid - 1) - pp.second.first,
                       make_pair(pp.second.first, mid - 1)));
    d.insert(make_pair(pp.second.second - (mid + 1),
                       make_pair(mid + 1, pp.second.second)));
  }
  for (long long i = 0; i < n; i++) cout << ans[i] << " ";
  cout << "\n";
  return;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long tc;
  cin >> tc;
  while (tc--) {
    long long n;
    cin >> n;
    if (n == 1) {
      cout << 1 << "\n";
      continue;
    } else {
      solver(n);
    }
  }
  return 0;
}

### 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;
const int MOD = 1e9 + 7;
struct cmp {
  bool operator()(const pair<long long, pair<long long, long long>> &a,
                  const pair<long long, pair<long long, long long>> &b) const {
    if (a.first != b.first)
      return a.first > b.first;
    else {
      return a.second.first < b.second.first;
    }
  }
};
void solver(long long n) {
  vector<long long> ans(n + 1, 0);
  set<pair<long long, pair<long long, long long>>, cmp> d;
  d.insert({n, {0, n - 1}});
  long long ctr = 1;
  while (d.size() != 0) {
    pair<long long, pair<long long, long long>> pp = *d.begin();
    d.erase(d.begin());
    if (pp.second.first < 0 || pp.second.second < 0 ||
        pp.second.first > pp.second.second)
      continue;
    long long mid = (pp.second.first + pp.second.second) / 2;
    ans[mid] = ctr++;
    d.insert(make_pair((mid - 1) - pp.second.first,
                       make_pair(pp.second.first, mid - 1)));
    d.insert(make_pair(pp.second.second - (mid + 1),
                       make_pair(mid + 1, pp.second.second)));
  }
  for (long long i = 0; i < n; i++) cout << ans[i] << " ";
  cout << "\n";
  return;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long tc;
  cin >> tc;
  while (tc--) {
    long long n;
    cin >> n;
    if (n == 1) {
      cout << 1 << "\n";
      continue;
    } else {
      solver(n);
    }
  }
  return 0;
}

```