#include <bits/stdc++.h>
using namespace std;
struct SteCompare {
  bool operator()(const pair<int, int> &p1, const pair<int, int> &p2) {
    int l1 = p1.second - p1.first + 1;
    int l2 = p2.second - p2.first + 1;
    if (l1 == l2) return p1.first < p2.first;
    return l1 > l2;
  }
};
template <typename TYPE>
void readint(TYPE &x) {
  x = 0;
  int f = 1;
  char c;
  for (c = getchar(); !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  x *= f;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int T;
  readint(T);
  while (T--) {
    int n;
    readint(n);
    int arr[n + 1], i = 1;
    priority_queue<pair<int, pair<int, int>>> pq;
    pq.push(make_pair(n, make_pair(-1, n)));
    while (!pq.empty()) {
      auto it = pq.top();
      pq.pop();
      int L = -it.second.first;
      int R = it.second.second;
      int mid = (L + R) / 2;
      arr[mid] = i++;
      if (L < mid) pq.push(make_pair(mid - L, make_pair(-L, mid - 1)));
      if (mid < R) pq.push(make_pair(R - mid, make_pair(-(mid + 1), R)));
    }
    for (i = 1; i <= n; i++) cout << arr[i] << " ";
    if (T) cout << '\n';
  }
  return 0;
}

### Prompt

Your task is to create 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;
struct SteCompare {
  bool operator()(const pair<int, int> &p1, const pair<int, int> &p2) {
    int l1 = p1.second - p1.first + 1;
    int l2 = p2.second - p2.first + 1;
    if (l1 == l2) return p1.first < p2.first;
    return l1 > l2;
  }
};
template <typename TYPE>
void readint(TYPE &x) {
  x = 0;
  int f = 1;
  char c;
  for (c = getchar(); !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  x *= f;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int T;
  readint(T);
  while (T--) {
    int n;
    readint(n);
    int arr[n + 1], i = 1;
    priority_queue<pair<int, pair<int, int>>> pq;
    pq.push(make_pair(n, make_pair(-1, n)));
    while (!pq.empty()) {
      auto it = pq.top();
      pq.pop();
      int L = -it.second.first;
      int R = it.second.second;
      int mid = (L + R) / 2;
      arr[mid] = i++;
      if (L < mid) pq.push(make_pair(mid - L, make_pair(-L, mid - 1)));
      if (mid < R) pq.push(make_pair(R - mid, make_pair(-(mid + 1), R)));
    }
    for (i = 1; i <= n; i++) cout << arr[i] << " ";
    if (T) cout << '\n';
  }
  return 0;
}

```