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 Comp {
  bool operator()(const pair<int, int> &a, const pair<int, int> &b) {
    int lenA = a.second - a.first;
    int lenB = b.second - b.first;
    if (lenA < lenB) {
      return true;
    } else if (lenA > lenB) {
      return false;
    } else {
      if (a.first > b.first) {
        return true;
      } else {
        return false;
      }
    }
  }
};
int main() {
  int T, n;
  cin >> T;
  while (T-- > 0) {
    cin >> n;
    vector<int> output(n, 0);
    priority_queue<pair<int, int>, vector<pair<int, int>>, Comp> pq;
    int counter = 1;
    pq.push({0, n - 1});
    while (!pq.empty()) {
      pair<int, int> top = pq.top();
      pq.pop();
      int a = top.first;
      int b = top.second;
      int mid = (a + b) / 2;
      output[mid] = counter++;
      if (mid - a > 0) {
        pq.push({a, mid - 1});
      }
      if (b - mid > 0) {
        pq.push({mid + 1, b});
      }
    }
    for (int i = 0; i < output.size(); i++) {
      cout << output[i];
      if (i != (output.size() - 1)) {
        cout << " ";
      }
    }
    cout << endl;
  }
  return 0;
}