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;
int t;
int n;
struct subarray {
  int size, l, r;
} f;
bool cmp(subarray a, subarray b) {
  if (a.size != b.size) {
    return a.size > b.size;
  }
  return a.l < b.l;
}
set<subarray, bool (*)(subarray, subarray)> st(cmp);
int l, r, curr, sz, v[200123], mid;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> t;
  while (t--) {
    cin >> n;
    st.insert({n, 1, n});
    curr = 1;
    for (int i = 1; i <= n; ++i) {
      v[i] = 0;
    }
    while (!st.empty()) {
      f = *st.begin(), st.erase(st.begin());
      l = f.l, r = f.r;
      if ((r - l + 1) % 2) {
        mid = (l + r) / 2;
        if (v[mid] == 0) v[mid] = curr++;
      } else {
        mid = (l + r - 1) / 2;
        if (v[mid] == 0) v[mid] = curr++;
      }
      if (l < r) {
        st.insert({mid - l, l, mid - 1});
        st.insert({r - mid, mid + 1, r});
      }
    }
    for (int i = 1; i <= n; ++i) {
      cout << v[i] << ' ';
    }
    cout << '\n';
  }
  return 0;
}