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;
#pragma comment(linker, "/STACK:1024000000,1024000000")
const long long mod = 1000000007;
const long long inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
long long qpow(long long a, long long b) {
  long long s = 1;
  while (b > 0) {
    if (b % 2 == 1) {
      s = s * a;
    }
    a = a * a;
    b = b >> 1;
  }
  return s;
}
long long qpowmod(long long a, long long b, long long c) {
  long long res, t;
  res = 1;
  t = a % c;
  while (b) {
    if (b & 1) {
      res = res * t % c;
    }
    t = t * t % c;
    b >>= 1;
  }
  return res;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long read() {
  long long k = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    k = k * 10 + ch - '0';
    ch = getchar();
  }
  return k * f;
}
inline void write(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
const long long maxn = 2e6 + 5;
long long t, n, a[maxn];
bool f(long long x, long long y) {
  long long ans = 0;
  for (long long i = x; i <= y; i++) {
    ans |= i;
  }
  cout << x << ' ' << y << ' ' << ans << ' ' << y - x + 1 << "\n";
  if (ans < y - x + 1)
    cout << x << ' ' << y << ' ' << ans << ' ' << y - x + 1 << "\n";
}
signed main() {
  cin >> t;
  while (t--) {
    cin >> n;
    for (long long i = 1; i <= n; i++) {
      cout << i << ' ';
    }
    cout << "\n";
  }
  return 0;
}
