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>
using namespace std;
const int PI = 3.141592653589793;
const int inf = 1000111222;
const int mod = 1000000007;
const int N = 200007;
bool comp(const pair<int, int> &a, const pair<int, int> &b) {
  int len = a.second - a.first + 1;
  int len2 = b.second - b.first + 1;
  if (len == len2) return a.first < b.first;
  return (len > len2);
}
void solve() {
  int n;
  cin >> n;
  int ar[n + 7];
  set<pair<int, int>, bool (*)(const pair<int, int> &, const pair<int, int> &)>
      S(&comp);
  S.insert({1, n});
  int cnt = 0;
  while (!S.empty()) {
    auto it = S.begin();
    int l = it->first, r = it->second;
    S.erase(it);
    int len = r - l + 1;
    int mid = 0;
    if (len & 1) {
      mid = (l + r) / 2;
    } else {
      mid = (l + r - 1) / 2;
    }
    ++cnt;
    ar[mid] = cnt;
    if (len == 1) continue;
    if (mid > l) S.insert({l, mid - 1});
    if (mid < r) S.insert({mid + 1, r});
  }
  for (int i = 1; i <= n; i++) {
    cout << ar[i] << " ";
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}