#include <bits/stdc++.h>
using namespace std;
long long expo(long long x, long long n, long long m) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return expo((x * x) % m, n / 2, m);
  else
    return (x * expo((x * x) % m, (n - 1) / 2, m)) % m;
}
string toBinary(long long n) {
  std::string r;
  while (n != 0) {
    r = (n % 2 == 0 ? "0" : "1") + r;
    n /= 2;
  }
  return r;
}
void solve() {
  long long n;
  cin >> n;
  long long ans[n + 1];
  priority_queue<pair<long long, pair<long long, long long> > > q;
  q.push({n, {n - 1, n}});
  long long k = 1;
  while (!q.empty()) {
    long long s = q.top().first;
    long long l = n - q.top().second.first;
    long long r = q.top().second.second;
    long long d = s;
    q.pop();
    if (s == 1) {
      ans[l] = k;
      k++;
      continue;
    }
    if (d % 2 == 0) {
      ans[(l + r - 1) / 2] = k;
      if ((l + r - 1) / 2 != l) q.push({d / 2 - 1, {n - l, (r + l) / 2 - 1}});
      q.push({d / 2, {n - ((r + l) / 2 + 1), r}});
    } else {
      ans[(r + l) / 2] = k;
      q.push({d / 2, {n - l, (l + r) / 2 - 1}});
      q.push({d / 2, {n - ((l + r) / 2 + 1), r}});
    }
    k++;
  }
  for (long long i = 1; i <= n; i++) cout << ans[i] << " ";
  cout << endl;
  return;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  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;
long long expo(long long x, long long n, long long m) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return expo((x * x) % m, n / 2, m);
  else
    return (x * expo((x * x) % m, (n - 1) / 2, m)) % m;
}
string toBinary(long long n) {
  std::string r;
  while (n != 0) {
    r = (n % 2 == 0 ? "0" : "1") + r;
    n /= 2;
  }
  return r;
}
void solve() {
  long long n;
  cin >> n;
  long long ans[n + 1];
  priority_queue<pair<long long, pair<long long, long long> > > q;
  q.push({n, {n - 1, n}});
  long long k = 1;
  while (!q.empty()) {
    long long s = q.top().first;
    long long l = n - q.top().second.first;
    long long r = q.top().second.second;
    long long d = s;
    q.pop();
    if (s == 1) {
      ans[l] = k;
      k++;
      continue;
    }
    if (d % 2 == 0) {
      ans[(l + r - 1) / 2] = k;
      if ((l + r - 1) / 2 != l) q.push({d / 2 - 1, {n - l, (r + l) / 2 - 1}});
      q.push({d / 2, {n - ((r + l) / 2 + 1), r}});
    } else {
      ans[(r + l) / 2] = k;
      q.push({d / 2, {n - l, (l + r) / 2 - 1}});
      q.push({d / 2, {n - ((l + r) / 2 + 1), r}});
    }
    k++;
  }
  for (long long i = 1; i <= n; i++) cout << ans[i] << " ";
  cout << endl;
  return;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```