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;
struct node {
  int l;
  int r;
  int len;
};
bool operator<(node a, node b) {
  if (b.len > a.len)
    return true;
  else if (b.len < a.len)
    return false;
  else {
    return b.l < a.l;
  }
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    priority_queue<node> q;
    int ans[n];
    node t1, t2, t3;
    t1.l = 0;
    t1.r = n - 1;
    t1.len = n;
    q.push(t1);
    int ct = 0;
    while (!q.empty()) {
      t2 = q.top();
      q.pop();
      ans[(t2.l + t2.r) / 2] = ct + 1;
      ct++;
      if (t2.l == t2.r) {
        continue;
      }
      if ((t2.r - t2.l + 1) % 2 == 0) {
        if (((t2.r + t2.l) / 2 + 1) <= t2.r) {
          t3.l = (t2.r + t2.l) / 2 + 1;
          t3.r = t2.r;
          t3.len = t3.r - t3.l + 1;
          q.push(t3);
        }
        if (((t2.l + t2.r) / 2 - 1) >= t2.l) {
          t3.l = t2.l;
          t3.r = (t2.l + t2.r) / 2 - 1;
          t3.len = t3.r - t3.l + 1;
          q.push(t3);
        }
      } else {
        if (((t2.l + t2.r) / 2 - 1) >= t2.l) {
          t3.l = t2.l;
          t3.r = (t2.l + t2.r) / 2 - 1;
          t3.len = t3.r - t3.l + 1;
          q.push(t3);
        }
        if (((t2.r + t2.l) / 2 + 1) <= t2.r) {
          t3.l = (t2.r + t2.l) / 2 + 1;
          t3.r = t2.r;
          t3.len = t3.r - t3.l + 1;
          q.push(t3);
        }
      }
    }
    for (int i = 0; i < n; ++i) {
      cout << ans[i] << " ";
    }
    cout << endl;
  }
}