#include <bits/stdc++.h>
using namespace std;
bool isvowel(char c) {
  if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' || c == 'A' ||
      c == 'E' || c == 'I' || c == 'O' || c == 'U')
    return true;
  return false;
}
long long fast_exp(long long x, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) ans *= x;
    n = n >> 1;
    x = x * x;
  }
  return ans;
}
long long mod_exp(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long sod(long long n) {
  long long sum = 0;
  while (n != 0) {
    sum = sum + (n % 10);
    n = n / 10;
  }
  return sum;
}
long long modInverse(long long a, long long m) { return mod_exp(a, m - 2, m); }
long long nCrModpDP(long long n, long long r, long long p) {
  long long C[r + 1];
  memset(C, 0, sizeof(C));
  C[0] = 1;
  for (long long i = 1; i <= n; i++) {
    for (long long j = min(i, r); j > 0; j--) C[j] = (C[j] + C[j - 1]) % p;
  }
  return C[r];
}
long long nCrModpLucas(long long n, long long r, long long p) {
  if (r == 0) return 1;
  long long ni = n % p, ri = r % p;
  return (nCrModpLucas(n / p, r / p, p) * nCrModpDP(ni, ri, p)) % p;
}
struct CustomComparator {
  bool operator()(pair<long long, pair<long long, long long>> x,
                  pair<long long, pair<long long, long long>> y) {
    if (x.first == y.first) return (x.second).first < (y.second).first;
    return x.first > y.first;
  }
};
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n, 0);
  long long cur = 1;
  set<pair<long long, pair<long long, long long>>, CustomComparator> st;
  st.insert({n, {0, n - 1}});
  while (!st.empty()) {
    pair<long long, pair<long long, long long>> bar = *(st.begin());
    st.erase(bar);
    long long x, y, idx;
    x = (bar.second).first;
    y = (bar.second).second;
    idx = (x + y) / 2;
    a[idx] = cur;
    if (x <= idx - 1) st.insert({idx - x, {x, idx - 1}});
    if (y >= idx + 1) st.insert({y - idx, {idx + 1, y}});
    cur += 1;
  }
  for (long long i = (long long)0; i < (long long)n; i++) cout << a[i] << " ";
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) solve();
}

### Prompt

Your task is to create a Cpp solution to the following 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 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
bool isvowel(char c) {
  if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' || c == 'A' ||
      c == 'E' || c == 'I' || c == 'O' || c == 'U')
    return true;
  return false;
}
long long fast_exp(long long x, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) ans *= x;
    n = n >> 1;
    x = x * x;
  }
  return ans;
}
long long mod_exp(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long sod(long long n) {
  long long sum = 0;
  while (n != 0) {
    sum = sum + (n % 10);
    n = n / 10;
  }
  return sum;
}
long long modInverse(long long a, long long m) { return mod_exp(a, m - 2, m); }
long long nCrModpDP(long long n, long long r, long long p) {
  long long C[r + 1];
  memset(C, 0, sizeof(C));
  C[0] = 1;
  for (long long i = 1; i <= n; i++) {
    for (long long j = min(i, r); j > 0; j--) C[j] = (C[j] + C[j - 1]) % p;
  }
  return C[r];
}
long long nCrModpLucas(long long n, long long r, long long p) {
  if (r == 0) return 1;
  long long ni = n % p, ri = r % p;
  return (nCrModpLucas(n / p, r / p, p) * nCrModpDP(ni, ri, p)) % p;
}
struct CustomComparator {
  bool operator()(pair<long long, pair<long long, long long>> x,
                  pair<long long, pair<long long, long long>> y) {
    if (x.first == y.first) return (x.second).first < (y.second).first;
    return x.first > y.first;
  }
};
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n, 0);
  long long cur = 1;
  set<pair<long long, pair<long long, long long>>, CustomComparator> st;
  st.insert({n, {0, n - 1}});
  while (!st.empty()) {
    pair<long long, pair<long long, long long>> bar = *(st.begin());
    st.erase(bar);
    long long x, y, idx;
    x = (bar.second).first;
    y = (bar.second).second;
    idx = (x + y) / 2;
    a[idx] = cur;
    if (x <= idx - 1) st.insert({idx - x, {x, idx - 1}});
    if (y >= idx + 1) st.insert({y - idx, {idx + 1, y}});
    cur += 1;
  }
  for (long long i = (long long)0; i < (long long)n; i++) cout << a[i] << " ";
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) solve();
}

```