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;
vector<pair<long long int, pair<long long int, long long int>>> v1;
long long int power(long long int x, long long int n) {
  long long int result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result * x);
    x = x * x;
    n = n / 2;
  }
  return result;
}
using abcd = pair<long long int, pair<long long int, long long int>>;
bool compare(abcd& a1, abcd& a2) {
  long long int s1 = a1.first;
  long long int s2 = a2.first;
  if (s1 != s2) return (s2 < s1);
  long long int p1 = a1.second.first;
  long long int p2 = a2.second.first;
  return (p1 < p2);
}
void fun(long long int start, long long int end) {
  if (start > end) return;
  long long int mid = (start + end) / 2;
  v1.push_back(make_pair(end - start + 1, make_pair(start, end)));
  fun(start, mid - 1);
  fun(mid + 1, end);
}
void solve() {
  long long int n;
  cin >> n;
  fun(0, n - 1);
  sort(v1.begin(), v1.begin() + n, compare);
  vector<long long int> v2(n);
  for (long long int i = 0; i <= n - 1; i++) {
    long long int a1 = (v1[i].second.first + v1[i].second.second) / 2;
    v2[a1] = i + 1;
  }
  for (long long int i = 0; i <= n - 1; i++) cout << v2[i] << " ";
  cout << endl;
  v1.clear();
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}