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;
long long int modp(long long int a, long long int p, long long int n) {
  long long int d[100];
  long long int i, j, k, l;
  if (n == 0) return 1;
  for (i = 0; n >= 1; i++) {
    d[i] = n % 2;
    n /= 2;
  }
  l = i;
  long long int e = 1;
  if (d[0] == 1) e *= (a % p);
  for (i = 1; i < l; i++) {
    a *= a;
    a = a % p;
    if (d[i] == 1) {
      e *= a;
      e = e % p;
    }
  }
  return e % p;
}
long long int modInverse(long long int n, long long int p) {
  return modp(n, p, p - 2);
}
long long int p1 = 1e8 + 7, p2 = 998244353;
long long int l, r, mid, ans = 0;
long long int n, i, j, k, g, m;
long long int x, y, n1, n2, h, z, c;
long long int aa = 1;
long long int ss = 1;
long long int maxSubArraySum(vector<long long int> &a, long long int size) {
  long long int max_so_far = INT_MIN, max_ending_here = 0;
  for (long long int i = 0; i < size; i++) {
    max_ending_here = max_ending_here + a[i];
    if (max_so_far < max_ending_here) max_so_far = max_ending_here;
    if (max_ending_here < 0) max_ending_here = 0;
  }
  return max_so_far;
}
void solve() {
  cin >> n;
  for (i = 0; i < n; i++) {
    cout << i + 1 << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  cin >> t;
  for (int i = 0; i < t; i++) {
    solve();
    cout << "\n";
  }
  return 0;
}
