#include <bits/stdc++.h>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
long long int x = 0, dp[200005];
class DConstructingTheArray {
 public:
  void solve(std::istream& in, std::ostream& out) {
    int t;
    in >> t;
    while (t-- > 0) {
      long long int n, i = 1;
      in >> n;
      set<long long int> st;
      map<long long int, set<long long int>> mp;
      long long int a[n + 1];
      memset(a, 0, sizeof(a));
      st.insert(n);
      mp[n].insert(1ll);
      while (!st.empty()) {
        long long int x = *st.rbegin();
        st.erase(x);
        for (auto p : mp[x]) {
          a[(x - 1) / 2 + p] = i;
          i++;
          if ((x - 1) / 2 > 0) {
            st.insert((x - 1) / 2);
            mp[(x - 1) / 2].insert(p);
          }
          if (x - (x - 1) / 2 - 1 > 0) {
            st.insert(x - (x - 1) / 2 - 1);
            mp[x - (x - 1) / 2 - 1].insert((x - 1) / 2 + p + 1);
          }
        }
        mp[x].clear();
      }
      for (i = 1; i <= n; i++) {
        out << a[i] << " ";
      }
      out << "\n";
    }
  }
};
int main() {
  DConstructingTheArray solver;
  std::ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}

### Prompt

Your challenge is to write 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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
long long int x = 0, dp[200005];
class DConstructingTheArray {
 public:
  void solve(std::istream& in, std::ostream& out) {
    int t;
    in >> t;
    while (t-- > 0) {
      long long int n, i = 1;
      in >> n;
      set<long long int> st;
      map<long long int, set<long long int>> mp;
      long long int a[n + 1];
      memset(a, 0, sizeof(a));
      st.insert(n);
      mp[n].insert(1ll);
      while (!st.empty()) {
        long long int x = *st.rbegin();
        st.erase(x);
        for (auto p : mp[x]) {
          a[(x - 1) / 2 + p] = i;
          i++;
          if ((x - 1) / 2 > 0) {
            st.insert((x - 1) / 2);
            mp[(x - 1) / 2].insert(p);
          }
          if (x - (x - 1) / 2 - 1 > 0) {
            st.insert(x - (x - 1) / 2 - 1);
            mp[x - (x - 1) / 2 - 1].insert((x - 1) / 2 + p + 1);
          }
        }
        mp[x].clear();
      }
      for (i = 1; i <= n; i++) {
        out << a[i] << " ";
      }
      out << "\n";
    }
  }
};
int main() {
  DConstructingTheArray solver;
  std::ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}

```