#include <bits/stdc++.h>
using namespace std;
int s[500500];
struct node {
  int l;
  int r;
};
struct cmp1 {
  bool operator()(const node &x, const node &y) const {
    if (x.r - x.l != y.r - y.l) {
      return (x.r - x.l) < (y.r - y.l);
    } else {
      return x.r > y.r;
    }
  }
};
void solve() {
  priority_queue<node, vector<node>, cmp1> q;
  int n;
  cin >> n;
  int k = 1;
  q.push((node){1, n});
  while (!q.empty()) {
    node ss = q.top();
    q.pop();
    int n = ss.r - ss.l + 1;
    if (n % 2 == 0) {
      int m = (ss.l + ss.r - 1) / 2;
      s[m] = k++;
      if (m - 1 >= ss.l) q.push((node){ss.l, m - 1});
      if (m + 1 <= ss.r) q.push((node){m + 1, ss.r});
    } else {
      int m = (ss.l + ss.r) / 2;
      s[m] = k++;
      if (m - 1 >= ss.l) q.push((node){ss.l, m - 1});
      if (m + 1 <= ss.r) q.push((node){m + 1, ss.r});
    }
  }
  for (int i = 1; i <= n; i++) {
    if (i == 1) {
      printf("%d", s[i]);
    } else {
      printf(" %d", s[i]);
    }
  }
  printf("\n");
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Generate a CPP solution to the following 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 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int s[500500];
struct node {
  int l;
  int r;
};
struct cmp1 {
  bool operator()(const node &x, const node &y) const {
    if (x.r - x.l != y.r - y.l) {
      return (x.r - x.l) < (y.r - y.l);
    } else {
      return x.r > y.r;
    }
  }
};
void solve() {
  priority_queue<node, vector<node>, cmp1> q;
  int n;
  cin >> n;
  int k = 1;
  q.push((node){1, n});
  while (!q.empty()) {
    node ss = q.top();
    q.pop();
    int n = ss.r - ss.l + 1;
    if (n % 2 == 0) {
      int m = (ss.l + ss.r - 1) / 2;
      s[m] = k++;
      if (m - 1 >= ss.l) q.push((node){ss.l, m - 1});
      if (m + 1 <= ss.r) q.push((node){m + 1, ss.r});
    } else {
      int m = (ss.l + ss.r) / 2;
      s[m] = k++;
      if (m - 1 >= ss.l) q.push((node){ss.l, m - 1});
      if (m + 1 <= ss.r) q.push((node){m + 1, ss.r});
    }
  }
  for (int i = 1; i <= n; i++) {
    if (i == 1) {
      printf("%d", s[i]);
    } else {
      printf(" %d", s[i]);
    }
  }
  printf("\n");
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```