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 pii {
  pii(int a, int b) : fi(a), se(b) {}
  int fi, se;
  bool operator<(const pii ot) const {
    return fi == ot.fi ? se > ot.se : fi < ot.fi;
  }
};
int ret[200005];
int T, N;
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &N);
    priority_queue<pii> pq;
    pq.push(pii(N, 1));
    int idx = 1;
    while (!pq.empty()) {
      int len = pq.top().fi;
      int st = pq.top().se;
      pq.pop();
      int m = (st + st + len - 1) >> 1;
      ret[m] = idx++;
      int rlen = len >> 1;
      int llen = len - rlen - 1;
      if (llen > 0) pq.push(pii(llen, st));
      if (rlen > 0) pq.push(pii(rlen, st + llen + 1));
    }
    for (int i = 1; i <= N; i++) printf("%d ", ret[i]);
    printf("\n");
  }
}