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;
using ll = long long;
const int N = 200005;
int ans[N];
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    priority_queue<pair<int, int>> q;
    q.emplace(n, 0);
    for (int val = 1; !q.empty(); val++) {
      int len, l;
      tie(len, l) = q.top();
      q.pop();
      l = -l;
      if (len % 2 == 1) {
        ans[l + len / 2] = val;
        len /= 2;
        if (len > 0) {
          q.emplace(len, -l);
          q.emplace(len, -(l + len + 1));
        }
      } else {
        ans[l + (len - 1) / 2] = val;
        len /= 2;
        if (len - 1 > 0) {
          q.emplace(len - 1, -l);
        }
        q.emplace(len, -(l + len));
      }
    }
    for (int i = 0; i < n; i++) {
      printf("%d%c", ans[i], " \n"[i + 1 == n]);
    }
  }
}