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 a[200010], k = 0, n;
struct node {
  int l, r;
} b[200010];
bool cmp(node x, node y) {
  int t1 = x.r - x.l, t2 = y.r - y.l;
  if (t1 == t2)
    return x.l < y.l;
  else
    return t1 > t2;
}
void bfs(int l, int r) {
  struct node u;
  u.l = l;
  u.r = r;
  queue<node> q;
  q.push(u);
  while (!q.empty()) {
    struct node u = q.front();
    q.pop();
    int l = u.l, r = u.r;
    b[++k].l = l, b[k].r = r;
    if ((r - l + 1) & 1) {
      int x = (l + r) / 2;
      if (x - 1 >= l) {
        struct node v;
        v.l = l;
        v.r = x - 1;
        q.push(v);
      }
      if (r >= x + 1) {
        struct node v;
        v.l = x + 1;
        v.r = r;
        q.push(v);
      }
    } else {
      int x = (l + r - 1) / 2;
      if (x - 1 >= l) {
        struct node v;
        v.l = l;
        v.r = x - 1;
        q.push(v);
      }
      if (r >= x + 1) {
        struct node v;
        v.l = x + 1;
        v.r = r;
        q.push(v);
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int T;
  cin >> T;
  while (T--) {
    cin >> n;
    k = 0;
    for (int i = 1; i <= n; ++i) a[i] = 0;
    bfs(1, n);
    sort(b + 1, b + k + 1, cmp);
    for (int i = 1; i <= k; ++i) {
      int l = b[i].l, r = b[i].r;
      if ((r - l + 1) & 1) {
        int x = (l + r) / 2;
        a[x] = i;
      } else {
        int x = (l + r - 1) / 2;
        a[x] = i;
      }
    }
    for (int i = 1; i <= n; ++i) {
      cout << a[i] << " ";
    }
    cout << endl;
  }
  return 0;
}