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;
long long n;
long long ncr(long long n, long long r) {
  long long i, ans = 1, k = 0;
  if (r == 0)
    return 1;
  else if (n < r)
    return 0;
  for (i = n; i >= n - r + 1; i--) {
    ans *= i;
    ans /= n - i + 1;
  }
  return ans;
}
void max_heapify(int a[100], int i) {
  int l, r, largest = i;
  l = 2 * i + 1;
  r = 2 * i + 2;
  if (l < n && a[l] > a[i]) largest = l;
  if (r < n && a[r] > a[l]) largest = r;
  if (largest != i) {
    int k;
    k = a[i];
    a[i] = a[largest];
    a[largest] = k;
    max_heapify(a, largest);
  }
}
int power(long long x, long long y, long long p) {
  int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
bool square(long long n) {
  double x;
  x = sqrt(n);
  if (x == floor(x))
    return true;
  else
    return false;
}
bool prime(long long n) {
  long long i;
  for (i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
long long modi(long long a, long long m) { return power(a, m - 2, m); }
void kmp(string s, long long p[]) {
  int n = (int)s.length();
  p[0] = 0;
  for (int i = 1; i < n; i++) {
    int j = p[i - 1];
    while (j > 0 && s[i] != s[j]) j = p[j - 1];
    if (s[i] == s[j]) j++;
    p[i] = j;
  }
}
void zfunc(string s, long long z[]) {
  long long n = (long long)s.length();
  z[0] = 0;
  for (long long i = 1, l = 0, r = 0; i < n; ++i) {
    if (i <= r) z[i] = min(r - i + 1, z[i - l]);
    while (i + z[i] < n && s[z[i]] == s[i + z[i]]) ++z[i];
    if (i + z[i] - 1 > r) l = i, r = i + z[i] - 1;
  }
}
struct pair_hash {
  size_t operator()(const pair<int, int> &p) const noexcept {
    return hash<long long>()((((long long)p.first) << 32) |
                             ((long long)p.second));
  }
};
bool palin(string s) {
  long long i, n;
  n = s.length();
  for (i = 0; i <= n / 2; i++) {
    if (s[i] != s[n - i - 1]) {
      return false;
    }
  }
  return true;
}
long long sum(long long n) {
  long long i;
  i = n;
  long long sum = 0;
  while (i > 0) {
    sum += i % 10;
    i = i / 10;
  }
  return sum;
}
struct Compare {
  bool operator()(const pair<long long, pair<long long, long long>> &fff,
                  const pair<long long, pair<long long, long long>> &bbb) {
    if (fff.first == bbb.first) return fff.second.first > bbb.second.first;
    return fff.first < bbb.first;
  }
};
long long nCr(long long n, long long r, long long p, long long fac[]) {
  if (r == 0) return 1;
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % p;
  return (fac[n] * modi(fac[r], p) % p * modi(fac[n - r], p) % p) % p;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long mod(string num, long long a) {
  long long res = 0;
  for (long long i = 0; i < num.length(); i++)
    res = (res * 10 + (long long)num[i] - '0') % a;
  return res;
}
long long parent[1000], sizer[1000];
long long findset(long long a) {
  if (a == parent[a]) {
    return a;
  }
  parent[a] = findset(parent[a]);
  return parent[a];
}
void unionset(long long a, long long b) {
  a = findset(a);
  b = findset(b);
  if (sizer[a] < sizer[b]) {
    swap(a, b);
  }
  sizer[b] += sizer[a];
  parent[b] = a;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long T = 1, i, j;
  cin >> T;
  while (T--) {
    long long n;
    cin >> n;
    vector<long long> ans(n);
    priority_queue<pair<long long, pair<long long, long long>>,
                   vector<pair<long long, pair<long long, long long>>>, Compare>
        q;
    q.push({n, {0, n - 1}});
    long long done = 1;
    while (!q.empty()) {
      pair<long long, pair<long long, long long>> temp = q.top();
      q.pop();
      long long start = temp.second.first;
      long long end = temp.second.second;
      long long mid = (start + end) / 2;
      ans[mid] = done;
      if (start < mid) q.push({mid - start, {start, mid - 1}});
      if (mid < end) q.push({end - mid, {mid + 1, end}});
      done++;
    }
    for (i = 0; i < n; i++) cout << ans[i] << " ";
    cout << "\n";
  }
  return 0;
}