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>
int a[1000001], b[1000001], c[1000001], n, e = 2, dist[1000001], len = 1, v;
int comp(int x, int y) {
  int i, j, k;
  if ((c[x] - b[x]) > (c[y] - b[y])) return 1;
  if ((c[x] - b[x]) < (c[y] - b[y])) return 0;
  if (c[x] < c[y]) return 1;
  return 0;
}
void insert(int k, int j, int l) {
  a[n++] = k;
  b[k] = j;
  c[k] = l;
  int i, z = n - 1, x;
  while (z != 0) {
    i = a[(z - 1) / 2];
    x = a[z];
    if (comp(x, i) == 1) {
      a[z] = i;
      a[(z - 1) / 2] = x;
      z = (z - 1) / 2;
    } else
      break;
  }
}
void remov() {
  int i, j, k, z = 0, x;
  k = b[a[0]] + c[a[0]];
  dist[k / 2] = len++;
  if (k != 2) {
    k = k / 2;
    if (b[a[0]] != k) insert(e++, b[a[0]], k - 1);
    insert(e++, k + 1, c[a[0]]);
  }
  b[a[0]] = 400001;
  c[a[0]] = 400001;
  a[0] = a[n - 1];
  n--;
  a[n] = v + 2;
  while (z < n) {
    k = a[z];
    i = a[2 * z + 1];
    j = a[2 * z + 2];
    if ((comp(k, i) == 1) && (comp(k, j) == 1)) break;
    if (comp(i, j) == 1) {
      a[z] = i;
      a[2 * z + 1] = k;
      z = 2 * z + 1;
    } else {
      a[z] = j;
      a[2 * z + 2] = k;
      z = 2 * z + 2;
    }
  }
}
int main() {
  int i, j, k, t;
  scanf("%d", &t);
  for (i = 0; i <= 400002; i++) {
    b[i] = 400001;
    a[i] = 400001;
    c[i] = 400001;
  }
  for (int x = 0; x < t; x++) {
    n = 0;
    scanf("%d", &v);
    e = 2;
    len = 1;
    for (i = 0; i <= 2 * v + 10; i++) {
      b[i] = 400001;
      a[i] = v + 2;
      c[i] = 400001;
    }
    j = v;
    insert(1, 1, v);
    while (j > 0) {
      remov();
      j--;
    }
    dist[v] = v;
    for (i = 1; i <= v; i++) printf("%d ", dist[i]);
    printf("\n");
  }
}