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 n, k, m, d, rem;
long long *arr, *b;
vector<long long> mul_d;
struct comp {
  bool operator()(pair<long long, long long> p1,
                  pair<long long, long long> p2) {
    long long len1 = p1.second - p1.first + 1;
    long long len2 = p2.second - p2.first + 1;
    if (len2 == len1) {
      return p1.first < p2.first;
    }
    return len1 > len2;
  }
};
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    cin >> n;
    long long val = 1;
    arr = new long long[n]();
    set<pair<long long, long long>, comp> pending_blancks;
    pending_blancks.insert(make_pair(0, n - 1));
    long long pos;
    while (pending_blancks.size() != 0) {
      pair<long long, long long> cur = *pending_blancks.begin();
      pending_blancks.erase(cur);
      long long len = cur.second - cur.first + 1;
      long long mid = (cur.second + cur.first) / 2;
      if (len % 2 == 0) {
        pos = mid;
        arr[pos] = val++;
        long long si = pos + 1;
        long long ei = cur.second;
        if (si >= 0 && ei < n && si <= ei)
          pending_blancks.insert(make_pair(si, ei));
        si = cur.first;
        ei = pos - 1;
        if (si >= 0 && ei < n && si <= ei)
          pending_blancks.insert(make_pair(si, ei));
      } else {
        pos = mid;
        arr[pos] = val++;
        long long si = cur.first;
        long long ei = pos - 1;
        if (si >= 0 && ei < n && si <= ei)
          pending_blancks.insert(make_pair(si, ei));
        si = pos + 1;
        ei = cur.second;
        if (si >= 0 && ei < n && si <= ei)
          pending_blancks.insert(make_pair(si, ei));
      }
    }
    for (long long i = 0; i < n; i++) {
      cout << arr[i] << " ";
    }
    cout << "\n";
  }
}