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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
struct Seg {
  int beg, end;
  bool operator<(const Seg& other) const {
    return std::make_pair(end - beg, other.beg) <
           std::make_pair(other.end - other.beg, beg);
  }
};
std::priority_queue<Seg> queue;
int ans[200005];
void solve() {
  int n;
  std::cin >> n;
  queue.push({0, n});
  int cnt = 1;
  while (!queue.empty()) {
    auto a = queue.top();
    queue.pop();
    if (a.beg == a.end) {
      break;
    }
    int m(0);
    if ((a.end - a.beg) % 2) {
      m = (a.beg + a.end) / 2;
    } else {
      m = (a.beg + a.end) / 2 - 1;
    }
    ans[m] = cnt++;
    queue.push({a.beg, m});
    queue.push({m + 1, a.end});
  }
  while (!queue.empty()) {
    queue.pop();
  }
  for (int i = 0; i < n; i++) {
    std::cout << ans[i] << ' ';
  }
  std::cout << '\n';
}
signed main() {
  std::ios_base::sync_with_stdio(false);
  std::cout.tie(0);
  std::cin.tie(0);
  int t(1);
  std::cin >> t;
  while (t--) {
    solve();
  }
}