#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
struct comparatr {
  bool operator()(pair<long long, long long> const &p1,
                  pair<long long, long long> const &p2) {
    long long l1 = abs(p1.first - p1.second) + 1;
    long long l2 = abs(p2.first - p2.second) + 1;
    if (l1 == l2) return p1.first < p2.first;
    return l1 > l2;
  }
};
vector<long long> solve(long long n) {
  vector<long long> ans(n + 5);
  set<pair<long long, long long>, comparatr> pq;
  pq.insert({1, n});
  long long i = 1;
  while (!pq.empty()) {
    pair<long long, long long> p = *pq.begin();
    long long l = p.first, r = p.second;
    pq.erase(pq.begin());
    long long mid = (l + r) / 2;
    ans[mid] = i++;
    if (l < mid) pq.insert({l, mid - 1});
    if (mid < r) pq.insert({mid + 1, r});
  }
  return ans;
}
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(false);
  long long test = 1, N, m, n, h, itr, x, y, i, j, k, z, e, f;
  cin >> test;
  while (test--) {
    cin >> n;
    vector<long long> ans = solve(n);
    for (i = 1; i <= n; i++) cout << ans[i] << " ";
    ans.clear();
    cout << "\n";
  }
  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;
const long long mod = 1e9 + 7;
struct comparatr {
  bool operator()(pair<long long, long long> const &p1,
                  pair<long long, long long> const &p2) {
    long long l1 = abs(p1.first - p1.second) + 1;
    long long l2 = abs(p2.first - p2.second) + 1;
    if (l1 == l2) return p1.first < p2.first;
    return l1 > l2;
  }
};
vector<long long> solve(long long n) {
  vector<long long> ans(n + 5);
  set<pair<long long, long long>, comparatr> pq;
  pq.insert({1, n});
  long long i = 1;
  while (!pq.empty()) {
    pair<long long, long long> p = *pq.begin();
    long long l = p.first, r = p.second;
    pq.erase(pq.begin());
    long long mid = (l + r) / 2;
    ans[mid] = i++;
    if (l < mid) pq.insert({l, mid - 1});
    if (mid < r) pq.insert({mid + 1, r});
  }
  return ans;
}
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(false);
  long long test = 1, N, m, n, h, itr, x, y, i, j, k, z, e, f;
  cin >> test;
  while (test--) {
    cin >> n;
    vector<long long> ans = solve(n);
    for (i = 1; i <= n; i++) cout << ans[i] << " ";
    ans.clear();
    cout << "\n";
  }
  return 0;
}

```