#include <bits/stdc++.h>
using namespace std;
using pii = pair<int, int>;
using ll = long long;
struct segment {
  int l, r;
  bool operator<(const segment& other) const {
    if (r - l == other.r - other.l) {
      return l < other.l;
    }
    return r - l > other.r - other.l;
  }
};
void Solve() {
  int n;
  cin >> n;
  vector<int> ans(n, 0);
  set<segment> s;
  s.insert({0, n - 1});
  int iter = 1;
  while (!s.empty()) {
    auto top = *s.begin();
    s.erase(s.begin());
    int mid = (top.l + top.r) / 2;
    ans[mid] = iter;
    int newr = mid - 1;
    int newl = mid + 1;
    if (top.l <= newr) {
      s.insert({top.l, newr});
    }
    if (newl <= top.r) {
      s.insert({newl, top.r});
    }
    iter++;
  }
  for (const auto& elem : ans) {
    printf("%d ", elem);
  }
  printf("\n");
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    Solve();
  }
  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;
using pii = pair<int, int>;
using ll = long long;
struct segment {
  int l, r;
  bool operator<(const segment& other) const {
    if (r - l == other.r - other.l) {
      return l < other.l;
    }
    return r - l > other.r - other.l;
  }
};
void Solve() {
  int n;
  cin >> n;
  vector<int> ans(n, 0);
  set<segment> s;
  s.insert({0, n - 1});
  int iter = 1;
  while (!s.empty()) {
    auto top = *s.begin();
    s.erase(s.begin());
    int mid = (top.l + top.r) / 2;
    ans[mid] = iter;
    int newr = mid - 1;
    int newl = mid + 1;
    if (top.l <= newr) {
      s.insert({top.l, newr});
    }
    if (newl <= top.r) {
      s.insert({newl, top.r});
    }
    iter++;
  }
  for (const auto& elem : ans) {
    printf("%d ", elem);
  }
  printf("\n");
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    Solve();
  }
  return 0;
}

```