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;
void swapll(long long *a, long long *b) {
  long long tmp = *a;
  *a = *b;
  *b = tmp;
}
void swapc(char *a, char *b) {
  char tmp = *a;
  *a = *b;
  *b = tmp;
}
void solve() {
  long long n;
  cin >> n;
  set<array<long long, 2>> s;
  s.insert({-n, 0});
  vector<long long> ans(n, 0);
  for (long long i = 1; i <= n; i++) {
    auto x = *s.begin();
    s.erase(s.begin());
    long long len = -x[0], pos = x[1];
    long long cen = pos + (len - 1) / 2;
    ans[cen] = i;
    s.insert({-(cen - pos), pos});
    s.insert({-(len - (cen - pos) - 1), cen + 1});
  }
  for (long long i = 0; i < ans.size(); i++)
    cout << ans[i] << (i + 1 < ans.size() ? ' ' : '\n');
  ;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long testcases;
  cin >> testcases;
  while (testcases--) {
    solve();
  }
  return 0;
}