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 inf = 0x3f3f3f3f;
const int maxn = 2e5 + 5;
const long long mod = 1e9 + 7;
struct node {
  int l, r, lg;
  friend bool operator<(const node &a, const node &b) {
    if (a.lg == b.lg) return a.l > b.l;
    return a.lg < b.lg;
  }
  friend bool operator>(const node &a, const node &b) {
    if (a.lg == b.lg) return a.l < b.l;
    return a.lg > b.lg;
  }
};
bool cmp(node a, node b) {
  if (a.lg == b.lg) return a.l < b.l;
  return a.lg > b.lg;
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    priority_queue<node> V;
    int n;
    cin >> n;
    int cnt = 1;
    int a[n + 5];
    memset(a, 0, sizeof(a));
    node tmp;
    tmp.l = 1, tmp.lg = n, tmp.r = n;
    V.push(tmp);
    int size = 1, sizetmp = 1;
    while (1) {
      size = sizetmp;
      sizetmp = 0;
      for (long long i = 0; i < size; ++i) {
        node linshi = V.top();
        V.pop();
        int ln = linshi.l, rn = linshi.r;
        int dif = rn - ln + 1, use;
        if (1 == dif % 2) {
          use = (ln + rn) / 2;
        } else {
          use = (ln + rn - 1) / 2;
        }
        if (a[use])
          continue;
        else
          a[use] = cnt++;
        if (use - 1 >= ln) {
          node nd;
          nd.l = ln;
          nd.r = use - 1;
          nd.lg = use - ln;
          V.push(nd);
          sizetmp++;
        }
        if (use + 1 <= rn) {
          node nd;
          nd.l = use + 1;
          nd.r = rn;
          nd.lg = rn - use;
          V.push(nd);
          sizetmp++;
        }
      }
      if (0 == sizetmp) break;
    }
    ++n;
    for (long long i = 1; i < n; ++i) cout << a[i] << " ";
    cout << endl;
  }
  return 0;
}