#include <bits/stdc++.h>
using namespace std;
struct cmp {
  bool operator()(const pair<int, int> &a, const pair<int, int> &b) const {
    int la = a.second - a.first;
    int lb = b.second - b.first;
    if (la == lb)
      return a.first < b.first;
    else
      return la > lb;
  }
};
void solve() {
  int n;
  cin >> n;
  set<pair<int, int>, cmp> st;
  vector<int> ans(n + 1);
  int cnt = 1;
  st.insert(make_pair(1, n));
  while (cnt < n) {
    pair<int, int> cur = *st.begin();
    st.erase(st.begin());
    int id = cur.second - cur.first;
    int nid;
    if (id & 1) {
      nid = (cur.first + cur.second - 1) / 2;
      ans[nid] = cnt;
    } else {
      nid = (cur.first + cur.second) / 2;
      ans[nid] = cnt;
    }
    if (cur.first < nid) st.insert(make_pair(cur.first, nid - 1));
    if (cur.second > nid) st.insert(make_pair(nid + 1, cur.second));
    cnt++;
  }
  for (auto i = ans.begin() + 1; i != ans.end(); i++) {
    if (*i == 0)
      cout << n << " ";
    else
      cout << *i << " ";
  }
  cout << "\n";
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### 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;
struct cmp {
  bool operator()(const pair<int, int> &a, const pair<int, int> &b) const {
    int la = a.second - a.first;
    int lb = b.second - b.first;
    if (la == lb)
      return a.first < b.first;
    else
      return la > lb;
  }
};
void solve() {
  int n;
  cin >> n;
  set<pair<int, int>, cmp> st;
  vector<int> ans(n + 1);
  int cnt = 1;
  st.insert(make_pair(1, n));
  while (cnt < n) {
    pair<int, int> cur = *st.begin();
    st.erase(st.begin());
    int id = cur.second - cur.first;
    int nid;
    if (id & 1) {
      nid = (cur.first + cur.second - 1) / 2;
      ans[nid] = cnt;
    } else {
      nid = (cur.first + cur.second) / 2;
      ans[nid] = cnt;
    }
    if (cur.first < nid) st.insert(make_pair(cur.first, nid - 1));
    if (cur.second > nid) st.insert(make_pair(nid + 1, cur.second));
    cnt++;
  }
  for (auto i = ans.begin() + 1; i != ans.end(); i++) {
    if (*i == 0)
      cout << n << " ";
    else
      cout << *i << " ";
  }
  cout << "\n";
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```