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>
struct Seg {
  int left;
  int right;
};
bool cmp(const Seg &p1, const Seg &p2) {
  if ((p1.right - p1.left) < (p2.right - p2.left)) {
    return false;
  } else if ((p1.right - p1.left) > (p2.right - p2.left)) {
    return true;
  } else {
    if (p1.right < p2.right) {
      return true;
    } else {
      return false;
    }
  }
}
int main() {
  int q;
  scanf("%d", &q);
  while (q--) {
    int n;
    int last_max = 1;
    scanf("%d", &n);
    std::queue<Seg> q;
    std::vector<Seg> vec;
    int arr[200000] = {
        0,
    };
    q.push(Seg{0, n - 1});
    vec.push_back(Seg{0, n - 1});
    while (q.size() != 0) {
      Seg elem = q.front();
      q.pop();
      if (elem.left >= elem.right) {
        continue;
      }
      if ((elem.right - elem.left + 1) % 2 == 1) {
        int mid = (elem.right + elem.left) / 2;
        if (elem.left <= mid - 1) {
          q.push(Seg{elem.left, mid - 1});
          vec.push_back(Seg{elem.left, mid - 1});
        }
        if (mid + 1 <= elem.right) {
          q.push(Seg{mid + 1, elem.right});
          vec.push_back(Seg{mid + 1, elem.right});
        }
      } else {
        int mid = (elem.right + elem.left - 1) / 2;
        if (elem.left <= mid - 1) {
          q.push(Seg{elem.left, mid - 1});
          vec.push_back(Seg{elem.left, mid - 1});
        }
        if (mid + 1 <= elem.right) {
          q.push(Seg{mid + 1, elem.right});
          vec.push_back(Seg{mid + 1, elem.right});
        }
      }
    }
    std::sort(vec.begin(), vec.begin() + n, cmp);
    for (int i = 0; i < n; i++) {
      int len = vec[i].right - vec[i].left + 1;
      if (len % 2 == 1) {
        int mid = (vec[i].right + vec[i].left) / 2;
        arr[mid] = i + 1;
      } else {
        int mid = (vec[i].right + vec[i].left - 1) / 2;
        arr[mid] = i + 1;
      }
    }
    for (int i = 0; i < n; i++) {
      printf("%d ", arr[i]);
    }
    printf("\n");
  }
  return 0;
}