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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
double pi = 2.0 * acos(0);
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout << fixed << setprecision(10);
}
long long int max2(long long int a, long long int b) {
  if (a > b) return a;
  return b;
}
long long int min2(long long int a, long long int b) {
  if (a < b) return a;
  return b;
}
long long int max3(long long int a, long long int b, long long int c) {
  return max2(a, max2(b, c));
}
long long int min3(long long int a, long long int b, long long int c) {
  return min2(a, min2(b, c));
}
struct node {
  long long int size;
  long long int l;
  long long int r;
};
struct comp {
  bool operator()(node const& p1, node const& p2) {
    if (p1.size < p2.size)
      return true;
    else if (p1.size == p2.size && p1.l > p2.l)
      return true;
    return false;
  }
};
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> arr(n + 1);
  priority_queue<node, vector<node>, comp> q;
  struct node temp;
  temp.size = n;
  temp.l = 0;
  temp.r = n - 1;
  q.push(temp);
  long long int l, r, mid;
  long long int cnt = 1;
  long long int lsize;
  long long int rsize;
  while (!q.empty()) {
    node p = q.top();
    q.pop();
    l = p.l;
    r = p.r;
    if ((r - l + 1) % 2 == 0)
      mid = (l + r - 1) / 2;
    else
      mid = (l + r) / 2;
    arr[mid] = cnt;
    cnt++;
    lsize = mid - l;
    rsize = r - mid;
    if (lsize > 0) {
      temp.size = lsize;
      temp.l = l;
      temp.r = mid - 1;
      q.push(temp);
    }
    if (rsize > 0) {
      temp.size = rsize;
      temp.l = mid + 1;
      temp.r = r;
      q.push(temp);
    }
  }
  for (long long int i = 0; i < n; i++) cout << arr[i] << " ";
  cout << endl;
}
signed main(void) {
  fast();
  long long int t;
  cin >> t;
  while (t--) solve();
}