#include <bits/stdc++.h>
using namespace std;
int a[200010], cnt, tot;
struct node {
  int left, right, mid, delta;
} b[200010];
int cmp(node a, node b) {
  return a.delta == b.delta ? a.left < b.left : a.delta > b.delta;
}
void dfs(int left, int right) {
  int mid;
  if (left > right) return;
  tot++, b[tot].left = left, b[tot].right = right,
         b[tot].mid = (left + right) / 2, b[tot].delta = right - left + 1,
         mid = (left + right) / 2;
  dfs(left, mid - 1), dfs(mid + 1, right);
}
int main() {
  int t, n, i;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    for (i = 1; i <= n; i++) a[i] = 0;
    for (i = 1; i <= n; i++) b[i].left = b[i].right = b[i].delta = 0;
    cnt = 0, tot = 0;
    dfs(1, n);
    sort(b + 1, b + 1 + tot, cmp);
    for (i = 1; i <= tot; i++) a[b[i].mid] = ++cnt;
    for (i = 1; i <= n; i++) printf("%d ", a[i]);
    printf("\n");
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int a[200010], cnt, tot;
struct node {
  int left, right, mid, delta;
} b[200010];
int cmp(node a, node b) {
  return a.delta == b.delta ? a.left < b.left : a.delta > b.delta;
}
void dfs(int left, int right) {
  int mid;
  if (left > right) return;
  tot++, b[tot].left = left, b[tot].right = right,
         b[tot].mid = (left + right) / 2, b[tot].delta = right - left + 1,
         mid = (left + right) / 2;
  dfs(left, mid - 1), dfs(mid + 1, right);
}
int main() {
  int t, n, i;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    for (i = 1; i <= n; i++) a[i] = 0;
    for (i = 1; i <= n; i++) b[i].left = b[i].right = b[i].delta = 0;
    cnt = 0, tot = 0;
    dfs(1, n);
    sort(b + 1, b + 1 + tot, cmp);
    for (i = 1; i <= tot; i++) a[b[i].mid] = ++cnt;
    for (i = 1; i <= n; i++) printf("%d ", a[i]);
    printf("\n");
  }
  return 0;
}

```