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;
long long cnt;
void dnc(long long* arr, long long l, long long r) {
  if (l > r) {
    return;
  }
  cout << l << " " << r << endl;
  long long x = (l + r) >> 1;
  arr[x] = ++cnt;
  if (x - 1 - l >= r - x - 1) {
    dnc(arr, l, x - 1);
    dnc(arr, x + 1, r);
  } else {
    dnc(arr, x + 1, r);
    dnc(arr, l, x - 1);
  }
}
struct st {
  bool operator()(const pair<long long, long long>& p1,
                  const pair<long long, long long>& p2) {
    if (p1.first != p2.first) {
      return p1.first < p2.first;
    } else {
      return p1.second > p2.second;
    }
  }
};
bool findll(long long x, long long y, long long& len, long long& l) {
  if (x > y) {
    return false;
  }
  len = (y - x + 1);
  l = x;
  return true;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long arr[n];
    priority_queue<pair<long long, long long>,
                   vector<pair<long long, long long> >, st>
        pq;
    pq.push(make_pair(n, 0));
    long long cnt = 1;
    while (!pq.empty()) {
      pair<long long, long long> p = pq.top();
      long long x = p.second;
      long long y = p.second + p.first - 1;
      pq.pop();
      long long ind = (x + y) / 2;
      arr[ind] = cnt++;
      long long len1, l1;
      if (findll(x, ind - 1, len1, l1)) {
        pq.push(make_pair(len1, l1));
      }
      long long len2, l2;
      if (findll(ind + 1, y, len2, l2)) {
        pq.push(make_pair(len2, l2));
      }
    }
    for (long long i = 0; i < n; i++) cout << arr[i] << " ";
    cout << endl;
  }
  return 0;
}