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;
bool pr[10000005] = {false};
vector<long long> prime;
long long max1(long long a, long long b) {
  if (a > b)
    return a;
  else
    return b;
}
long long min1(long long a, long long b) {
  if (a < b)
    return a;
  else
    return b;
}
long long mul(long long a, long long b) {
  return ((a % 1000000007) * (b % 1000000007)) % 1000000007;
}
void prim() {
  pr[0] = true;
  pr[1] = true;
  for (long long i = 2; i <= sqrt(100005); i++) {
    for (long long j = i * 2; j <= sqrt(100005); j += i) {
      pr[j] = true;
    }
  }
}
vector<long long> divisor(long long n) {
  vector<long long> div;
  div.push_back(n);
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      div.push_back(i);
      if (n / i != i) div.push_back(n / i);
    }
  }
  sort(div.begin(), div.end());
  return div;
}
long long modInverse(long long a, long long m) {
  long long m0 = m;
  long long y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long q = a / m;
    long long t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += m0;
  return x;
}
long long find(long long n) {
  long long count = 0;
  while (n) {
    count++;
    n = n / 2;
  }
  return count;
}
void solve() {
  long long n;
  cin >> n;
  for (long long i = 0; i < (long long)n; i++) {
    cout << i + 1 << " ";
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
