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;
struct kek {
  int l, r;
  kek(int x = 0, int y = 0) { l = x, r = y; }
};
const bool operator<(const kek a, const kek b) {
  int aa = a.r - a.l, bb = b.r - b.l;
  if (aa != bb) {
    return aa > bb;
  }
  return a.l < b.l;
}
void solve() {
  int n;
  cin >> n;
  vector<int> ans(n);
  set<kek> setik;
  setik.insert(kek(0, n - 1));
  for (int i = 1; i <= n; ++i) {
    int l = setik.begin()->l, r = setik.begin()->r;
    setik.erase(setik.begin());
    int m = (l + r) / 2;
    ans[m] = i;
    setik.insert(kek(l, m - 1)), setik.insert(kek(m + 1, r));
  }
  for (int i = 0; i < n; ++i) {
    cout << ans[i] << ' ';
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
    cout << '\n';
  }
  return 0;
}