A permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).

For a positive integer n, we call a permutation p of length n good if the following condition holds for every pair i and j (1 ≤ i ≤ j ≤ n) — 

  * (p_i  OR  p_{i+1}  OR  …  OR  p_{j-1}  OR  p_{j}) ≥ j-i+1, where OR denotes the [bitwise OR operation.](https://en.wikipedia.org/wiki/Bitwise_operation#OR)



In other words, a permutation p is good if for every subarray of p, the OR of all elements in it is not less than the number of elements in that subarray. 

Given a positive integer n, output any good permutation of length n. We can show that for the given constraints such a permutation always exists.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.

The first and only line of every test case contains a single integer n (1 ≤ n ≤ 100).

Output

For every test, output any good permutation of length n on a separate line. 

Example

Input


3
1
3
7


Output


1
3 1 2
4 3 5 2 7 1 6

Note

For n = 3, [3,1,2] is a good permutation. Some of the subarrays are listed below. 

  * 3 OR 1 = 3 ≥ 2 (i = 1,j = 2) 
  * 3 OR 1 OR 2 = 3 ≥ 3 (i = 1,j = 3) 
  * 1 OR 2 = 3 ≥ 2 (i = 2,j = 3) 
  * 1 ≥ 1 (i = 2,j = 2) 



Similarly, you can verify that [4,3,5,2,7,1,6] is also good.
#include <bits/stdc++.h>
using namespace std;
void prt(long x) { cerr << x; }
void prt(long long x) { cerr << x; }
void prt(unsigned x) { cerr << x; }
void prt(unsigned long x) { cerr << x; }
void prt(unsigned long long x) { cerr << x; }
void prt(float x) { cerr << x; }
void prt(double x) { cerr << x; }
void prt(long double x) { cerr << x; }
void prt(char x) { cerr << '\'' << x << '\''; }
void prt(const char *x) { cerr << '\"' << x << '\"'; }
void prt(const string &x) { cerr << '\"' << x << '\"'; }
void prt(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void prt(const pair<T, V> &x) {
  cerr << '{';
  prt(x.first);
  cerr << ',';
  prt(x.second);
  cerr << '}';
}
template <typename T>
void prt(const T &x) {
  long long f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), prt(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  prt(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
void err(istream_iterator<string> it) {}
template <typename T, typename... brgs>
void err(istream_iterator<string> it, T a, brgs... args) {
  cerr << *it << " = " << a << "\n";
  err(++it, args...);
}
const long long mod = 1000000007;
const long long N = 3e5;
char alphabet[26] = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i',
                     'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r',
                     's', 't', 'u', 'v', 'w', 'x', 'y', 'z'};
map<long long, long long> Fib;
long long kadane(long long *a, long long n);
bool isPrime(long long n);
long long mpow(long long a, long long b);
long long fib(long long n);
long long cmp(long long a, long long b) { return a > b; }
void solve() {
  long long n;
  cin >> n;
  for (__typeof(n + 1) i = (1); i != (n + 1); i += 1 - 2 * ((1) > (n + 1)))
    cout << i << " ";
  cout << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) solve();
}
long long mpow(long long a, long long b) {
  long long m = mod;
  long long x = 1;
  while (b > 0) {
    if (b & 1) {
      x = (x * a) % m;
    }
    a = (a * a) % m;
    b >>= 1;
  }
  return x;
}
long long fib(long long n) {
  Fib[0] = Fib[1] = 1;
  if (Fib.count(n)) return Fib[n];
  long long k = n / 2;
  if (n % 2 == 0) {
    return Fib[n] = (fib(k) * fib(k) + fib(k - 1) * fib(k - 1)) % mod;
  } else {
    return Fib[n] = (fib(k) * fib(k + 1) + fib(k - 1) * fib(k)) % mod;
  }
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
long long kadane(long long *a, long long n) {
  long long max_so_far = INT_MIN, max_ending_here = 0;
  for (__typeof(n) i = (0); i != (n); i += 1 - 2 * ((0) > (n))) {
    max_ending_here += a[i];
    max_so_far = max(max_so_far, max_ending_here);
    if (max_ending_here < 0) max_ending_here = 0;
  }
  return max_so_far;
}
