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;
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
inline void normal(long long int &a) {
  a %= 1000000007;
  (a < 0) && (a += 1000000007);
}
inline long long int modMul(long long int a, long long int b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a * b) % 1000000007;
}
inline long long int modAdd(long long int a, long long int b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a + b) % 1000000007;
}
inline long long int modSub(long long int a, long long int b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline long long int modPow(long long int b, long long int p) {
  long long int r = 1;
  while (p) {
    if (p & 1) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1;
  }
  return r;
}
inline long long int modInverse(long long int a) {
  return modPow(a, 1000000007 - 2);
}
inline long long int modDiv(long long int a, long long int b) {
  return modMul(a, modInverse(b));
}
inline bool isInside(pair<int, int> p, long long int n, long long int m) {
  return (p.first >= 0 && p.first < n && p.second >= 0 && p.second < m);
}
inline bool isInside(pair<int, int> p, long long int n) {
  return (p.first >= 0 && p.first < n && p.second >= 0 && p.second < n);
}
inline bool isSquare(long long int x) {
  long long int s = sqrt(x);
  return (s * s == x);
}
inline bool isFib(long long int x) {
  return isSquare(5 * x * x + 4) || isSquare(5 * x * x - 4);
}
inline bool isPowerOfTwo(long long int x) {
  return ((1LL << (long long int)log2(x)) == x);
}
template <typename first, typename second>
ostream &operator<<(ostream &os, const pair<first, second> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "{";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename first, typename second>
ostream &operator<<(ostream &os, const map<first, second> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
clock_t tStart = clock();
void faltu() { cerr << endl; }
template <typename T>
void faltu(T a[], int n) {
  for (int i = 0; i < n; ++i) cerr << a[i] << ' ';
  cerr << endl;
}
template <typename T, typename... hello>
void faltu(T arg, const hello &...rest) {
  cerr << arg << ' ';
  faltu(rest...);
}
struct func {
  bool operator()(pair<int, int> const &a, pair<int, int> const &b) {
    if (a.first == b.first) return (a.second < b.second);
    return (a.first < b.first);
  }
};
const long long int INF = 0x3f3f3f3f3f3f3f3f;
const long double EPS = 1e-9;
const int inf = 0x3f3f3f3f;
const int mx = (int)1e5 + 9;
long long int n, m, a, b, t, i, j, d, cs = 0, counT = 0, k, ans = 0, l = 0,
                                      sum1 = 0, sum = 0, Max, Min, num;
vector<long long int> vc;
map<long long int, long long int> mp;
int main() {
  ios_base::sync_with_stdio(false);
  ;
  t = 1;
  cin >> t;
  while (t--) {
    cin >> n;
    for (i = n; i >= 1; i--) cout << i << ' ';
    cout << '\n';
  }
}
