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;
typedef struct _node {
  int left;
  int right;
  int length;
  _node(int left, int right)
      : left(left), right(right), length(right - left + 1) {}
} Node;
struct cmp {
  bool operator()(Node n1, Node n2) {
    if (n1.length < n2.length) {
      return true;
    } else if (n1.length == n2.length) {
      if (n1.left > n2.left) {
        return true;
      }
    }
    return false;
  }
};
int f(int* arr, int left, int right, int idx) {
  int mid = -1;
  if ((right - left + 1) % 2 == 1) {
    arr[(left + right) / 2] = idx;
    mid = (left + right) / 2;
  } else {
    arr[(left + right - 1) / 2] = idx;
    mid = (left + right - 1) / 2;
  }
  return mid;
}
int main() {
  int t;
  scanf("%d", &t);
  for (int tc = 0; tc < t; tc++) {
    int n;
    scanf("%d", &n);
    int arr[200002] = {
        0,
    };
    priority_queue<Node, vector<Node>, cmp> pq;
    pq.push(Node(1, n));
    int idx = 0;
    while (1) {
      idx++;
      if (pq.empty()) {
        break;
      }
      Node temp = pq.top();
      pq.pop();
      int tLeft = temp.left;
      int tRight = temp.right;
      int mid = f(arr, tLeft, tRight, idx);
      if (mid - 1 >= tLeft) {
        pq.push(Node(tLeft, mid - 1));
      }
      if (mid + 1 <= tRight) {
        pq.push(Node(mid + 1, tRight));
      }
    }
    for (int i = 1; i <= n; i++) {
      printf("%d ", arr[i]);
    }
    printf("\n");
  }
}