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 kN = 2e5 + 5;
int mid, idx, t, n, a[kN];
struct node {
  int l, r;
  friend bool operator<(node a, node b) {
    if ((a.r - a.l) == (b.r - b.l)) return a.l > b.l;
    return (a.r - a.l) < (b.r - b.l);
  }
} cur;
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  priority_queue<node> q;
  while (t--) {
    cin >> n;
    cur.l = 1, cur.r = n;
    q.push(cur);
    for (int i = 1; i <= n; i++) {
      cur = q.top();
      q.pop();
      idx = cur.r - cur.l + 1;
      if (cur.r == cur.l) {
        a[cur.r] = i;
        continue;
      }
      if (idx & 1) {
        mid = cur.r + cur.l >> 1;
        a[mid] = i;
      } else {
        mid = cur.l + cur.r - 1 >> 1;
        a[mid] = i;
      }
      q.push({cur.l, mid - 1});
      q.push({mid + 1, cur.r});
    }
    for (int i = 1; i <= n; i++) {
      if (i != 1) cout << ' ';
      cout << a[i];
    }
    cout << '\n';
    while (!q.empty()) q.pop();
  }
  return 0;
}