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;
double tick() {
  static clock_t oldt, newt = clock();
  double diff = 1.0 * (newt - oldt) / CLOCKS_PER_SEC;
  oldt = newt;
  return diff;
}
vector<pair<int, int> > p;
void med(vector<int> &a, int l, int r, int cnt) {
  if (l < r) {
    p.push_back({l, r});
    int mid;
    if ((r - l + 1) % 2)
      mid = (l + r) / 2;
    else
      mid = (l + r - 1) / 2;
    a[mid] = cnt;
    int m1 = mid - l;
    int m2 = r - mid;
    if (m1 >= m2) {
      med(a, l, mid - 1, cnt + 1);
      med(a, mid + 1, r, cnt + 2);
    } else {
      med(a, mid + 1, r, cnt + 2);
      med(a, l, mid - 1, cnt + 1);
    }
  }
}
bool cmp(const pair<int, int> &a, const pair<int, int> &b) {
  int l1 = a.second - a.first + 1;
  int l2 = b.second - b.first + 1;
  if (l1 == l2) return (a.first < b.first);
  return l1 > l2;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    p.clear();
    int n, val = 0;
    cin >> n;
    vector<int> a(n + 1, 0);
    med(a, 1, n, 1);
    sort(p.begin(), p.end(), cmp);
    int cnt = 1;
    for (int i = 0; i < p.size(); i++, cnt++) {
      int x = p[i].first;
      int y = p[i].second;
      if ((y - x + 1) % 2) {
        int mid = (x + y) / 2;
        a[mid] = cnt;
      } else {
        int mid = (x + y - 1) / 2;
        a[mid] = cnt;
      }
    }
    for (int i = 1; i <= n; i++) val = max(val, a[i]);
    val += 1;
    for (int i = 1; i <= n; i++) {
      if (a[i] == 0) {
        a[i] = val;
        val++;
      }
    }
    for (int i = 1; i <= n; i++) cout << a[i] << " ";
    cout << "\n";
  }
  cerr << "execution time is: " << tick() << "\n";
}