#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pi = pair<int, int>;
using pll = pair<long, long>;
using vi = vector<int>;
using vll = vector<ll>;
using vpi = vector<pi>;
using vpll = vector<pll>;
const int mod = 1e9 + 7;
const ll LLINF = 0x3f3f3f3f3f3f3f3f;
int v[200005];
int n;
struct comp_interv {
  bool operator()(pi a, pi b) {
    return (a.second - a.first + 1 < b.second - b.first + 1) ||
           (a.second - a.first + 1 == b.second - b.first + 1 &&
            a.first > b.first);
  }
};
void bfs() {
  priority_queue<pair<int, int>, vector<pi>, comp_interv> q;
  q.push({1, n});
  int cont = 1;
  while (!q.empty()) {
    auto curr = q.top();
    q.pop();
    int len = curr.second - curr.first + 1;
    int m = len % 2 == 0 ? (curr.first + curr.second - 1) / 2
                         : (curr.first + curr.second) / 2;
    v[m] = cont++;
    if (len == 1) continue;
    if (curr.first < m) q.push({curr.first, m - 1});
    if (curr.second > m) q.push({m + 1, curr.second});
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    memset(v, 0, sizeof(v));
    cin >> n;
    bfs();
    for (int i = int((1)); i <= int((n)); i++) cout << v[i] << ' ';
    cout << '\n';
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
using ll = long long;
using pi = pair<int, int>;
using pll = pair<long, long>;
using vi = vector<int>;
using vll = vector<ll>;
using vpi = vector<pi>;
using vpll = vector<pll>;
const int mod = 1e9 + 7;
const ll LLINF = 0x3f3f3f3f3f3f3f3f;
int v[200005];
int n;
struct comp_interv {
  bool operator()(pi a, pi b) {
    return (a.second - a.first + 1 < b.second - b.first + 1) ||
           (a.second - a.first + 1 == b.second - b.first + 1 &&
            a.first > b.first);
  }
};
void bfs() {
  priority_queue<pair<int, int>, vector<pi>, comp_interv> q;
  q.push({1, n});
  int cont = 1;
  while (!q.empty()) {
    auto curr = q.top();
    q.pop();
    int len = curr.second - curr.first + 1;
    int m = len % 2 == 0 ? (curr.first + curr.second - 1) / 2
                         : (curr.first + curr.second) / 2;
    v[m] = cont++;
    if (len == 1) continue;
    if (curr.first < m) q.push({curr.first, m - 1});
    if (curr.second > m) q.push({m + 1, curr.second});
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    memset(v, 0, sizeof(v));
    cin >> n;
    bfs();
    for (int i = int((1)); i <= int((n)); i++) cout << v[i] << ' ';
    cout << '\n';
  }
  return 0;
}

```