#include <bits/stdc++.h>
using namespace std;
long long int power(long long int a, long long int b) {
  long long int res = 1;
  a = a % 1000000007;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % 1000000007;
      b--;
    }
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  return (b == 0) ? a : gcd(b, a % b);
}
struct cmp {
  bool operator()(const pair<long long int, long long int>& a,
                  const pair<long long int, long long int>& b) const {
    long long int len1 = a.second - a.first + 1;
    long long int len2 = b.second - b.first + 1;
    if (len1 != len2) return len1 > len2;
    return a.first < b.first;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    set<pair<long long int, long long int>, cmp> s;
    vector<long long int> v(n, 0);
    s.insert({0, n - 1});
    long long int c = 1;
    while (c <= n) {
      pair<long long int, long long int> cur = *s.begin();
      s.erase(s.begin());
      long long int l = cur.first;
      long long int r = cur.second;
      long long int len = r - l + 1;
      if (len % 2) {
        v[(l + r) / 2] = c++;
        if (l != r) {
          long long int mid = (l + r) / 2;
          s.insert({l, mid - 1});
          s.insert({mid + 1, r});
        }
      } else {
        v[(l + r - 1) / 2] = c++;
        long long int mid = (l + r - 1) / 2;
        s.insert({l, mid - 1});
        s.insert({mid + 1, r});
      }
    }
    for (long long int i = 0; i < n; i++) cout << v[i] << " ";
    cout << "\n";
  }
}

### 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;
long long int power(long long int a, long long int b) {
  long long int res = 1;
  a = a % 1000000007;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % 1000000007;
      b--;
    }
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  return (b == 0) ? a : gcd(b, a % b);
}
struct cmp {
  bool operator()(const pair<long long int, long long int>& a,
                  const pair<long long int, long long int>& b) const {
    long long int len1 = a.second - a.first + 1;
    long long int len2 = b.second - b.first + 1;
    if (len1 != len2) return len1 > len2;
    return a.first < b.first;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    set<pair<long long int, long long int>, cmp> s;
    vector<long long int> v(n, 0);
    s.insert({0, n - 1});
    long long int c = 1;
    while (c <= n) {
      pair<long long int, long long int> cur = *s.begin();
      s.erase(s.begin());
      long long int l = cur.first;
      long long int r = cur.second;
      long long int len = r - l + 1;
      if (len % 2) {
        v[(l + r) / 2] = c++;
        if (l != r) {
          long long int mid = (l + r) / 2;
          s.insert({l, mid - 1});
          s.insert({mid + 1, r});
        }
      } else {
        v[(l + r - 1) / 2] = c++;
        long long int mid = (l + r - 1) / 2;
        s.insert({l, mid - 1});
        s.insert({mid + 1, r});
      }
    }
    for (long long int i = 0; i < n; i++) cout << v[i] << " ";
    cout << "\n";
  }
}

```