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<long long int> isprime;
void seive(long long int size) {
  isprime.resize(size + 1, 1);
  long long int i, j;
  isprime[0] = 0;
  isprime[1] = 0;
  for (i = 2; i * i <= size; i++) {
    if (isprime[i]) {
      for (j = 2; j * i <= size; j++) isprime[i * j] = 0;
    }
  }
}
long long int power(long long int x, unsigned long long int y,
                    long long int m) {
  if (y == 0) return 1;
  long long int p = power(x, y / 2, m);
  p = (p * p) % m;
  if (y % 2 != 0) p = (p * x) % m;
  return p;
}
long long int modinverse(long long int a, long long int m) {
  return power(a, m - 2, m);
}
bool cmp(pair<long long int, long long int> a,
         pair<long long int, long long int> b) {
  return ((a.first > b.first) || (a.first == b.first && a.second < b.second));
}
long long int findans() {
  long long int n, i;
  cin >> n;
  vector<long long int> v(n, -2);
  queue<pair<long long int, long long int> > q;
  q.push({0, n - 1});
  long long int cnt = 1;
  long long int rem = n;
  vector<pair<long long int, long long int> > vt;
  while (!q.empty()) {
    pair<long long int, long long int> p = q.front();
    long long int l = p.first;
    long long int r = p.second;
    q.pop();
    long long int len = r - l + 1;
    if (len % 2 == 0) {
      long long int mid = (l + r - 1) / 2;
      if (mid < r) q.push({mid + 1, r});
      if (l < mid) q.push({l, mid - 1});
      vt.push_back({r - l + 1, mid});
      rem--;
    } else {
      long long int mid = (l + r) / 2;
      if (l < mid) q.push({l, mid - 1});
      if (mid < r) q.push({mid + 1, r});
      vt.push_back({r - l + 1, mid});
      rem--;
    }
  }
  sort((vt.begin()), (vt.end()), cmp);
  for (i = 0; i < n; i++) {
    v[vt[i].second] = cnt++;
  }
  for (i = 0; i < n; i++) cout << v[i] << " ";
  cout << endl;
  return 0;
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t, i;
  cin >> t;
  for (i = 0; i < t; i++) {
    findans();
  }
  return 0;
}