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 MXN = (int)2e5 + 7;
struct kek {
  int l, r, len;
  kek() {}
  kek(int l, int r, int len) : l(l), r(r), len(len) {}
};
bool cmp(kek &a, kek &b) {
  if (a.len == b.len) {
    return a.l < b.l;
  }
  return a.len > b.len;
}
int n, a[MXN], cnt, T;
vector<kek> node;
void comp(int l, int r) {
  int mid = (((r - l + 1) & 1) == 1 ? (r + l) / 2 : (r + l - 1) / 2);
  if (r - l + 1 > 1) {
    node.push_back(kek(l, r, r - l + 1));
  }
  if (r - l + 1 <= 2) return;
  comp(l, mid - 1);
  comp(mid + 1, r);
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    cin >> n;
    comp(1, n);
    sort(node.begin(), node.end(), cmp);
    for (kek i : node) {
      int mid = ((i.len & 1) == 1 ? (i.r + i.l) / 2 : (i.l + i.r - 1) / 2);
      a[mid] = ++cnt;
    }
    for (int i = 1; i <= n; i++) {
      if (!a[i]) a[i] = ++cnt;
      cout << a[i] << ' ';
      a[i] = 0;
    }
    cout << endl;
    cnt = 0;
    node.clear();
  }
  return 0;
}