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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline long long int GCD(long long int x, long long int y) {
  if (x < y) swap(x, y);
  if (x == 0) return y;
  if (y == 0) return x;
  return GCD(x % y, y);
}
long long int phi(long long int n) {
  long long int result = n;
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      while (n % i == 0) n /= i;
      result -= result / i;
    }
  }
  if (n > 1) result -= result / n;
  return result;
}
long long int power(long long int x, long long int n, long long int mod) {
  long long int res = 1;
  x %= mod;
  while (n) {
    if (n & 1) {
      res = ((res * x) % mod + mod) % mod;
    }
    x = ((x * x) % mod + mod);
    n >>= 1;
  }
  return res;
}
struct compareH {
  bool operator()(
      const pair<long long int, pair<long long int, long long int> > &a,
      const pair<long long int, pair<long long int, long long int> > &b) {
    if (a.first < b.first) return true;
    if (a.first == b.first) {
      if (a.second.first >= b.second.first) return true;
      return false;
    }
    return false;
  }
};
bool myfun(const pair<long long int, pair<long long int, long long int> > &a,
           const pair<long long int, pair<long long int, long long int> > &b) {
  if (a.first < b.first) return true;
  if (a.first == b.first) {
    if (a.second.first >= b.second.first) return true;
    return false;
  }
  return false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    priority_queue<
        pair<long long int, pair<long long int, long long int> >,
        vector<pair<long long int, pair<long long int, long long int> > >,
        compareH>
        q;
    q.push(make_pair(n, make_pair(1, n)));
    pair<long long int, pair<long long int, long long int> > dum;
    long long int i = 1;
    vector<long long int> A(n + 1);
    while (!q.empty()) {
      dum = q.top();
      q.pop();
      long long int len = dum.first;
      long long int fi = dum.second.first;
      long long int se = dum.second.second;
      if (fi > se) continue;
      long long int mid;
      if (len % 2) {
        mid = (se + fi) / 2;
        A[mid] = i++;
        q.push(make_pair(len / 2, make_pair(fi, mid - 1)));
        q.push(make_pair(len / 2, make_pair(mid + 1, se)));
      } else {
        mid = (se + fi - 1) / 2;
        A[mid] = i++;
        q.push(make_pair(len / 2, make_pair(mid + 1, se)));
        q.push(make_pair((len / 2) - 1, make_pair(fi, mid - 1)));
      }
    }
    for (long long int i = 1; i <= n; i++) {
      cout << A[i] << " ";
    }
    cout << endl;
  }
}