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;
template <class T>
void printvec(vector<T>& x, int a, int b) {
  for (int i = (a); i <= (b); i++) cout << x[i] << " ";
  cout << "\n";
}
const double PI = acos((double)-1);
const double EPSILON = 1e-7;
const int N = 1e6 + 1;
const int NN = 1e3 + 1;
const long long MOD = 1e9 + 7;
const long long oo = LLONG_MAX;
const int BASE = 10000;
void query() {
  int n;
  cin >> n;
  vector<int> a(n + 2);
  auto cmp = [](pair<int, int> a, pair<int, int> b) {
    return (a.second - a.first < b.second - b.first) ||
           (a.second - a.first == b.second - b.first && a.first > b.first);
  };
  priority_queue<pair<int, int>, vector<pair<int, int> >, decltype(cmp)> heap(
      cmp);
  heap.push({1, n});
  int time = 1;
  while (!heap.empty()) {
    pair<int, int> cur = heap.top();
    heap.pop();
    int mid = (cur.first + cur.second) / 2;
    a[mid] = time++;
    if (mid > cur.first) {
      heap.push({cur.first, mid - 1});
    }
    if (cur.second > mid) {
      heap.push({mid + 1, cur.second});
    }
  }
  printvec<int>(a, 1, n);
}
void solve() {
  int q;
  cin >> q;
  while (q--) {
    query();
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout << fixed << setprecision(10);
  solve();
  cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
  return 0;
}