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;
void file() {
  freopen("max.in", "r", stdin);
  freopen("max.out", "w", stdout);
}
const long long MOD = (long long)1e9 + 7ll;
const long long mod = 998244353;
const long long T = (long long)1e18;
const long long K = (long long)1e9;
const long long N = 3e5 + 100;
const long long tk = 1 << 17;
long long binpow(long long a, long long b) {
  long long ans = 1, k = a;
  while (b) {
    if (b % 2) ans *= k;
    k *= k;
    b /= 2;
  }
  return ans;
}
long long a[N];
long long n, c, r, l, x;
void solve() {
  for (long long i = 1; i <= n; i++) cout << a[i] << " ";
  cout << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long tek;
  cin >> tek;
  while (tek--) {
    cin >> n;
    set<pair<long long, pair<long long, long long>>> st;
    st.insert(make_pair(-n, make_pair(1, n)));
    c = 1;
    while (st.size()) {
      pair<long long, pair<long long, long long>> temp = *st.begin();
      r = temp.second.second;
      l = temp.second.first;
      st.erase(temp);
      if (l - r > 0) continue;
      if (l - r == 0) {
        a[l] = c++;
        continue;
      }
      x = r - l + 1;
      if (!(x % 2)) {
        a[(l + r - 1) / 2] = c++;
        x = (l + r - 1) / 2;
        st.insert(make_pair(-(x - l), make_pair(l, x - 1)));
        st.insert(make_pair(-(r - x), make_pair(x + 1, r)));
      } else {
        a[(l + r) / 2] = c++;
        x = (l + r) / 2;
        st.insert(make_pair(-(x - l), make_pair(l, x - 1)));
        st.insert(make_pair(-(r - x), make_pair(x + 1, r)));
      }
    }
    solve();
  }
  return 0;
}