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;
int a[210000];
int n;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    priority_queue<pair<int, int> > q;
    q.push({n - 1, -1});
    int x = 1;
    while (!q.empty()) {
      int l, r, d, e;
      d = q.top().first;
      e = q.top().second * -1;
      l = e;
      r = e + d;
      q.pop();
      if ((r - l + 1) % 2 == 1) {
        int y = (l + r) / 2;
        a[y] = x;
        x++;
        int sz;
        if (y - l >= r - y) {
          if (l <= y - 1) q.push({y - 1 - l, -l});
          if (y + 1 <= r) q.push({r - (y + 1), -(y + 1)});
        } else {
          if (y + 1 <= r) q.push({r - (y + 1), -(y + 1)});
          if (l <= y - 1) q.push({y - l - 1, -l});
        }
      } else {
        int y = (l + r - 1) / 2;
        a[y] = x;
        x++;
        if (y - l >= r - y) {
          if (l <= y - 1) q.push({y - 1 - l, -l});
          if (y + 1 <= r) q.push({r - (y + 1), -(y + 1)});
        } else {
          if (y + 1 <= r) q.push({r - (y + 1), -(y + 1)});
          if (l <= y - 1) q.push({y - l - 1, -l});
        }
      }
    }
    for (int i = 1; i < n + 1; i++) {
      printf("%d ", a[i]);
    }
    cout << endl;
  }
}