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;
struct pr {
  long long int first, second;
};
struct cmp {
  bool operator()(pr p1, pr p2) {
    if (p1.second - p1.first < p2.second - p2.first) return true;
    if (p1.second - p1.first == p2.second - p2.first && p1.first > p2.first)
      return true;
    return false;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    vector<long long int> v(n + 1);
    priority_queue<pr, vector<pr>, cmp> pq;
    long long int k = 0;
    pr p1;
    p1.first = 1, p1.second = n;
    pq.push(p1);
    while (!pq.empty()) {
      pr p = pq.top();
      pq.pop();
      long long int x = p.first;
      long long int y = p.second;
      if ((y - x + 1) % 2 == 1) {
        v[(x + y) / 2] = ++k;
        p1.first = x, p1.second = (x + y) / 2 - 1;
        if (x <= (x + y) / 2 - 1) pq.push(p1);
        p1.first = (x + y) / 2 + 1, p1.second = y;
        if ((x + y) / 2 + 1 <= y) pq.push(p1);
      } else {
        v[(x + y - 1) / 2] = ++k;
        p1.first = x, p1.second = (x + y - 1) / 2 - 1;
        if (x <= (x + y - 1) / 2 - 1) pq.push({x, (x + y - 1) / 2 - 1});
        p1.first = (x + y - 1) / 2 + 1, p1.second = y;
        if ((x + y - 1) / 2 + 1 <= y) pq.push({(x + y - 1) / 2 + 1, y});
      }
    }
    for (long long int i = 1; i <= n; i++) cout << v[i] << " ";
    cout << endl;
  }
  return 0;
}