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;
const int mN = 200200;
int r[mN], id[mN];
pair<int, int> elems[mN];
int median(int lo, int hi) {
  int cnt = hi - lo + 1;
  return (cnt & 1) ? (lo + hi) / 2 : (lo + hi - 1) / 2;
}
struct TSegment {
  int lo, hi, len;
  TSegment(int l = 0, int r = 0) : lo(l), hi(r), len(r - l) {}
  bool operator<(TSegment other) const {
    return !(len == other.len ? lo < other.lo : len > other.len);
  }
};
void giveRanks(int n) {
  priority_queue<TSegment, vector<TSegment>> heap;
  heap.push(TSegment(1, n));
  int rnk = 0;
  while (heap.size()) {
    TSegment seq = heap.top();
    heap.pop();
    int pos = median(seq.lo, seq.hi);
    r[pos] = rnk++;
    if (pos + 1 <= seq.hi) heap.push(TSegment(pos + 1, seq.hi));
    if (seq.lo <= pos - 1) heap.push(TSegment(seq.lo, pos - 1));
  }
}
bool cmp(pair<int, int> x, pair<int, int> y) {
  return (x.first == y.first) ? x.second < y.second : x.first < y.first;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    fill(r, r + 1 + n, 0);
    giveRanks(n);
    for (int i = 1; i <= n; i++) elems[i] = {r[i], i};
    sort(elems + 1, elems + 1 + n, cmp);
    for (int i = 1; i <= n; i++) id[elems[i].second] = i;
    for (int i = 1; i <= n; i++) cout << id[i] << " ";
    cout << "\n";
  }
}