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>
class duan {
 public:
  int l, r;
  friend bool operator<(duan a, duan b) {
    if (a.r - a.l < b.r - b.l) {
      return 1;
    } else if (a.r - a.l == b.r - b.l && a.r > b.r) {
      return 1;
    }
    return 0;
  }
};
int n;
int res[200005];
using namespace std;
void bfs(duan s) {
  int co = 1;
  priority_queue<duan> q;
  q.push(s);
  while (!q.empty()) {
    s = q.top();
    q.pop();
    int mid;
    if ((s.r - s.l + 1) % 2 == 1) {
      mid = (s.l + s.r) / 2;
    } else {
      mid = (s.l + s.r - 1) / 2;
    }
    res[mid] = co;
    co++;
    int l = mid - 1, r = mid + 1;
    if ((s.l - s.r + 1) % 2 == 0) {
      if (r < s.r) {
        duan temp;
        temp.l = r;
        temp.r = s.r;
        q.push(temp);
      }
      if (l > s.l) {
        duan temp;
        temp.l = s.l;
        temp.r = l;
        q.push(temp);
      }
    } else {
      if (l > s.l) {
        duan temp;
        temp.l = s.l;
        temp.r = l;
        q.push(temp);
      }
      if (r < s.r) {
        duan temp;
        temp.l = r;
        temp.r = s.r;
        q.push(temp);
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    if (res[i] == 0) {
      res[i] = co;
      co++;
    }
  }
}
int main() {
  int T = 1;
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      res[i] = 0;
    }
    if (n == 1) {
      printf("1\n");
      continue;
    } else if (n == 2) {
      printf("1 2\n");
      continue;
    }
    duan temp;
    temp.l = 1;
    temp.r = n;
    bfs(temp);
    for (int i = 1; i <= n; i++) {
      if (i == 1) {
        printf("%d", res[i]);
      } else {
        printf(" %d", res[i]);
      }
    }
    printf("\n");
  }
  return 0;
}