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;
const int N = 200010;
struct node {
  int x, y, len, pos;
  bool operator<(const node a) const {
    if (len != a.len)
      return len < a.len;
    else
      return pos > a.pos;
  }
};
int a[N];
priority_queue<node> q;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    memset(a, 0, sizeof a);
    scanf("%d", &n);
    q.push({1, n, n, 1});
    int k = 1;
    while (q.size()) {
      node t = q.top();
      q.pop();
      int l = t.x;
      int r = t.y;
      if (l == r) {
        continue;
      }
      int mid = (l + r) / 2;
      a[mid] = k++;
      int g1 = max(mid - 1, l);
      int g2 = min(mid + 1, r);
      q.push({l, g1, g1 - l + 1, l});
      q.push({g2, r, r - g2 + 1, g2});
    }
    for (int i = 1; i <= n; i++)
      if (a[i] == 0) a[i] = k++;
    for (int i = 1; i <= n; i++) cout << a[i] << " ";
    cout << endl;
  }
  return 0;
}