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;
void printVec(vector<int> v) {
  for (int i = 1; i < v.size(); i++) {
    cout << v[i] << " ";
  }
  cout << "\n";
}
void printMat(vector<vector<int> > mat) {
  for (int i = 0; i < mat.size(); i++) {
    for (int j = 0; j < mat[0].size(); j++) {
      cout << mat[i][j] << " ";
    }
    cout << endl;
  }
}
bool veccomp(pair<int, int> p1, pair<int, int> p2) {
  int t1 = (p1.second - p1.first + 1);
  int t2 = (p2.second - p2.first + 1);
  if (t1 < t2) {
    return true;
  } else if (t1 == t2) {
    if (p1.first < p2.first) {
      return false;
    } else {
      return true;
    }
  } else {
    return false;
  }
}
int main() {
  int T;
  cin >> T;
  while (T-- > 0) {
    int n;
    cin >> n;
    vector<int> res;
    res.assign(n + 1, 0);
    int count = 0;
    vector<pair<int, int> > q;
    q.push_back(pair<int, int>(1, n));
    pair<int, int> curr;
    int l, r;
    while (count < n) {
      count++;
      curr = q.front();
      pop_heap(q.begin(), q.end(), veccomp);
      q.pop_back();
      l = curr.first;
      r = curr.second;
      if ((r - l + 1) % 2 == 0) {
        res[(r + l - 1) / 2] = count;
        if (r >= (l + r - 1) / 2 + 1) {
          q.push_back(pair<int, int>((l + r - 1) / 2 + 1, r));
          push_heap(q.begin(), q.end(), veccomp);
        }
        if (l <= (l + r - 1) / 2 - 1) {
          q.push_back(pair<int, int>(l, (l + r - 1) / 2 - 1));
          push_heap(q.begin(), q.end(), veccomp);
        }
      } else {
        res[(r + l) / 2] = count;
        if (l <= (r + l) / 2 - 1) {
          q.push_back(pair<int, int>(l, (r + l) / 2 - 1));
          push_heap(q.begin(), q.end(), veccomp);
        }
        if (r >= (r + l) / 2 + 1) {
          q.push_back(pair<int, int>((r + l) / 2 + 1, r));
          push_heap(q.begin(), q.end(), veccomp);
        }
      }
    }
    printVec(res);
  }
}