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;
long long int mod = LLONG_MAX;
long long int power(long long int a, long long int b) {
  if (b == 0) return 1;
  long long int ans = power(a, b / 2);
  ans = (ans * ans) % mod;
  if (b % 2) ans *= a;
  return ans % mod;
}
long long int mul(long long int a, long long int n, long long int mod) {
  long long int ret;
  if (n == 0) return 0;
  ret = mul(a, n / 2, mod);
  ret = (ret + ret) % mod;
  if (n % 2) ret = (ret + a) % mod;
  return ret;
}
class comparator {
 public:
  bool operator()(const pair<int, pair<int, int> > a,
                  const pair<int, pair<int, int> > b) {
    if (a.first < b.first) return true;
    if (a.first > b.first) return false;
    if (a.second > b.second) return true;
    return false;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cout.tie(NULL);
  cin.tie(NULL);
  int t, n;
  cin >> t;
  while (t--) {
    cin >> n;
    vector<int> ans(n + 1, 0);
    int l, r, idx = 1;
    priority_queue<pair<int, pair<int, int> >,
                   vector<pair<int, pair<int, int> > >, comparator>
        pq;
    pq.push({n, {1, n}});
    while (!pq.empty()) {
      l = pq.top().second.first;
      r = pq.top().second.second;
      pq.pop();
      if (r < l) continue;
      if ((r - l + 1) % 2) {
        ans[(l + r) / 2] = idx++;
        if (r == l) continue;
        pq.push({(l + r) / 2 - l, {l, (l + r) / 2 - 1}});
        pq.push({r - (l + r) / 2, {(l + r) / 2 + 1, r}});
      } else {
        ans[(l + r - 1) / 2] = idx++;
        pq.push({(l + r - 1) / 2 - l, {l, (l + r - 1) / 2 - 1}});
        pq.push({r - (l + r - 1) / 2, {(l + r - 1) / 2 + 1, r}});
      }
    }
    for (int i = 1; i <= n; i++) cout << ans[i] << " ";
    cout << "\n";
  }
}