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 double pi = 3.14159265358979323;
long long int mod = 1000000000 + 7;
const double pii = acos(-1.0);
const long long int INF = 1e18;
const long long int inf = 1e9;
long long power(long long x, long long y) {
  long long res = 1;
  x = x;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
long long powe(long long x, long long y) {
  x = x % mod, y = y % (mod - 1);
  long long ans = 1;
  while (y > 0) {
    if (y & 1) {
      ans = (1ll * x * ans) % mod;
    }
    y >>= 1;
    x = (1ll * x * x) % mod;
  }
  return ans;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long lcm(long long a, long long b) { return (a / gcd(a, b) * b); }
bool isPrime(long long int n) {
  if (n < 2) return false;
  for (long long int i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
long long ncr(long long n, long long r) {
  long long res = 1;
  if (r > n) return 0;
  if (r > n - r) r = n - r;
  for (long long i = 0; i < r; i++) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
long long add(long long a, long long b) { return ((a % mod + b % mod) % mod); }
long long sub(long long a, long long b) {
  return ((a % mod - b % mod + mod) % mod);
}
long long mul(long long a, long long b) {
  return (((a % mod) * (b % mod)) % mod);
}
long long divi(long long a, long long b) {
  return (mul(a, powe(b, mod - 2)) % mod);
}
void fun() {}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  fun();
  long long int tests = 1;
  cin >> tests;
  while (tests--) {
    long long int n;
    cin >> n;
    priority_queue<pair<long long int, pair<long long int, long long int> > >
        pq;
    pq.push(make_pair(n, make_pair(-1, -n)));
    long long int a[n + 1];
    memset(a, 0, sizeof(a));
    for (long long int i = 1; i <= n; i++) {
      auto temp = pq.top();
      pq.pop();
      long long int len = temp.first;
      long long int l = -temp.second.first, r = -temp.second.second;
      long long int mid;
      if (len % 2)
        mid = (l + r) / 2;
      else
        mid = (l + r - 1) / 2;
      a[mid] = i;
      if (mid != l) pq.push(make_pair((mid - l), make_pair(-l, -(mid - 1))));
      if (mid != r) pq.push(make_pair((r - mid), make_pair(-(mid + 1), -r)));
    }
    for (long long int i = 1; i <= n; i++) cout << a[i] << ' ';
    cout << '\n';
  }
  cerr << "time elapsed : " << 1.0 * clock() / CLOCKS_PER_SEC << " sec \n";
  return 0;
}