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;
using ll = long long;
const int inf = INT_MAX;
const int ninf = INT_MIN;
const int mod = 1e9 + 7;
const int N = 1e6 + 2;
using pii = pair<int, int>;
int main() {
  int tt;
  scanf("%d", &tt);
  while (tt--) {
    int n;
    scanf("%d", &n);
    vector<int> a(n + 1, 0);
    priority_queue<pair<int, pii>> pq;
    pq.push(make_pair(n - 1, make_pair(-1, -n)));
    int nxt = 1;
    while (!pq.empty()) {
      pair<int, pii> here = pq.top();
      pq.pop();
      int gap = here.first, i = -here.second.first, j = -here.second.second;
      if (gap == 1) break;
      if ((j - i + 1) & 1) {
        int id = (j + i) / 2;
        a[id] = nxt++;
        int gap1 = id - i;
        if (id != i) {
          int gap1 = id - i;
          pq.push(make_pair(gap1, make_pair(-i, -(id - 1))));
        }
        if (id != j) {
          int gap2 = j - id;
          pq.push(make_pair(gap2, make_pair(-(id + 1), -j)));
        }
      } else {
        int id = (j + i - 1) / 2;
        a[id] = nxt++;
        if (id != i) {
          int gap1 = id - i;
          pq.push(make_pair(gap1, make_pair(-i, -(id - 1))));
        }
        if (id != j) {
          int gap2 = j - id;
          pq.push(make_pair(gap2, make_pair(-(id + 1), -j)));
        }
      }
    }
    for (int i = 1; i < (int)n + 1; i++)
      if (a[i] == 0) a[i] = nxt++;
    for (int i = 1; i < (int)n + 1; i++) printf("%d ", (int)a[i]);
    puts("");
  }
  return 0;
}