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;
const int maxn = 1000006, maxm = 1005;
const long long mod = 1000000007;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long powermod(long long a, long long b, long long c) {
  a = a % c;
  long long ans = 1;
  while (b) {
    if (b % 2) ans = (ans * a) % c;
    b = b / 2;
    a = (a * a) % c;
  }
  return ans;
}
long long ans[200005];
long long anss[200005];
set<long long> s;
int main() {
  long long T;
  set<long long>::iterator it;
  scanf("%lld", &T);
  while (T--) {
    s.clear();
    long long n, m, i, j, k;
    scanf("%lld", &n);
    for (i = 0; i <= n + 1; i++) ans[i] = 0;
    if (n == 1)
      printf("1\n");
    else {
      ans[(n + 1) / 2] = 1;
      s.insert((n + 1) / 2);
      long long x = (n + 1) / 2;
      long long cnt = 1;
      while (x) {
        if (cnt == n) break;
        long long vis = 0;
        long long x2 = x + 1;
        for (i = 1; i <= n; i++) {
          if (ans[i] == 0) {
            it = s.lower_bound(i);
            if (*it - i == x2 || (it == s.end() && n - i + 1 == x2)) {
              long long mid;
              if (*it - i == x2)
                mid = (i + *it - 1) / 2;
              else
                mid = (i + n) / 2;
              cnt++;
              ans[mid] = cnt;
              s.insert(mid);
              if (cnt == n) break;
            }
          }
        }
        if (cnt == n) break;
        x2 = x;
        for (i = 1; i <= n; i++) {
          if (ans[i] == 0) {
            it = s.lower_bound(i);
            if (*it - i == x2 || (it == s.end() && n - i + 1 == x2)) {
              long long mid;
              if (*it - i == x2)
                mid = (i + *it - 1) / 2;
              else
                mid = (i + n) / 2;
              cnt++;
              ans[mid] = cnt;
              s.insert(mid);
              if (cnt == n) break;
            }
          }
        }
        if (cnt == n) break;
        x2 = x - 1;
        for (i = 1; i <= n; i++) {
          if (ans[i] == 0) {
            it = s.lower_bound(i);
            if (*it - i == x2 || (it == s.end() && n - i + 1 == x2)) {
              long long mid;
              if (*it - i == x2)
                mid = (i + *it - 1) / 2;
              else
                mid = (i + n) / 2;
              cnt++;
              ans[mid] = cnt;
              s.insert(mid);
              if (cnt == n) break;
            }
          }
        }
        if (cnt == n) break;
        x = x / 2;
      }
      for (i = 1; i <= n; i++) printf("%lld ", ans[i]);
      printf("\n");
    }
  }
}