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 = (int)2e5 + 123;
const int K = (int)500;
const int maxn = (int)2e9;
const int mod = (int)1e9;
const long long inf = 1ll * 1e18 + 123;
int n;
int b[N], t;
set<pair<int, pair<int, int> > > s;
void clr() {
  for (int i = 0; i <= n; ++i) b[i] = 0;
}
void solve() {
  cin >> n;
  s.insert({-n, {1, n}});
  clr();
  t = 0;
  while (!s.empty()) {
    ++t;
    pair<int, pair<int, int> > a = *s.begin();
    int sz = a.first, l = a.second.first, r = a.second.second;
    sz *= -1;
    int id = (sz + 1) / 2;
    b[l + id - 1] = t;
    if (id > 1 && b[l + (id - 2)] == 0) {
      s.insert({-((l + (id - 2)) - l + 1), {l, l + id - 2}});
    }
    if (l + id <= r && b[l + id] == 0) {
      s.insert({-(r - (l + id) + 1), {l + id, r}});
    }
    s.erase(s.begin());
  }
  for (int i = 1; i <= n; ++i) {
    cout << b[i] << ' ';
  }
  cout << '\n';
}
int main() {
  ios_base ::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int q;
  q = 1;
  cin >> q;
  for (int i = 1; i <= q; ++i) {
    solve();
  }
  return 0;
}