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 long long N = 105;
char en = '\n';
long long inf = 2e16;
long long mod = 1e9 + 7;
long long power(long long x, long long n, long long mod) {
  long long res = 1;
  x %= mod;
  while (n) {
    if (n & 1) res = (res * x) % mod;
    x = (x * x) % mod;
    n >>= 1;
  }
  return res;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    map<pair<long long, long long>, pair<long long, long long>> que;
    long long res[n + 5];
    memset(res, 0, sizeof(res));
    que[{-n, 1}] = {1, n};
    long long oprn = 1;
    while (!que.empty()) {
      pair<long long, long long> curr = que.begin()->second;
      que.erase(que.begin());
      long long L = curr.first;
      long long R = curr.second;
      res[(L + R) / 2] = oprn;
      oprn++;
      if (L == R) continue;
      long long L1 = L, R1 = ((L + R) / 2) - 1;
      long long L2 = 1 + (L + R) / 2, R2 = R;
      if ((L2 - R2) > (L1 - R1)) {
        swap(L1, L2);
        swap(R1, R2);
      }
      if (L1 <= R1) que[{-(R1 - L1 + 1), L1}] = {L1, R1};
      if (L2 <= R2) que[{-(R2 - L2 + 1), L2}] = {L2, R2};
    }
    for (long long i = 1; i <= n; i++) cout << res[i] << " ";
    cout << en;
  }
  return 0;
}