description: Petya has an array of integers a_1, a_2, …, a_n. He only likes sorted arrays. Unfortunately, the given array could be arbitrary, so Petya wants to sort it.
  
  Petya likes to challenge himself, so he wants to sort array using only 3-cycles. More formally, in one operation he can pick 3 pairwise distinct indices i, j, and k (1 ≤ i, j, k ≤ n) and apply i → j → k → i cycle to the array a. It simultaneously places a_i on position j, a_j on position k, and a_k on position i, without changing any other element.
  
  For example, if a is [10, 50, 20, 30, 40, 60] and he chooses i = 2, j = 1, k = 5, then the array becomes [\underline{50}, \underline{40}, 20, 30, \underline{10}, 60].
  
  Petya can apply arbitrary number of 3-cycles (possibly, zero). You are to determine if Petya can sort his array a, i. e. make it non-decreasing.
  
  Input
  
  Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 5 ⋅ 10^5). Description of the test cases follows.
  
  The first line of each test case contains a single integer n (1 ≤ n ≤ 5 ⋅ 10^5) — the length of the array a.
  
  The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n).
  
  It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^5.
  
  Output
  
  For each test case, print "YES" (without quotes) if Petya can sort the array a using 3-cycles, and "NO" (without quotes) otherwise. You can print each letter in any case (upper or lower).
  
  Example
  
  Input
  
  
  7
  1
  1
  2
  2 2
  2
  2 1
  3
  1 2 3
  3
  2 1 3
  3
  3 1 2
  4
  2 1 4 3
  
  
  Output
  
  
  YES
  YES
  NO
  YES
  NO
  YES
  YES
  
  Note
  
  In the 6-th test case Petya can use the 3-cycle 1 → 3 → 2 → 1 to sort the array.
  
  In the 7-th test case Petya can apply 1 → 3 → 2 → 1 and make a = [1, 4, 2, 3]. Then he can apply 2 → 4 → 3 → 2 and finally sort the array.
solution: #include <bits/stdc++.h>
  using namespace std;
  void _print(bool t) { cerr << t; }
  void _print(long long t) { cerr << t; }
  void _print(string t) { cerr << t; }
  void _print(char t) { cerr << t; }
  void _print(double t) { cerr << t; }
  template <class T>
  void _print(vector<T> v) {
    cerr << "[  ";
    for (T i : v) {
      _print(i);
      cerr << "  ";
    }
    cerr << "]\n";
  }
  template <class T, class V>
  void _print(pair<T, V> p) {
    cerr << "{";
    _print(p.first);
    cerr << ", ";
    _print(p.second);
    cerr << "}";
  }
  template <class T>
  void _print(set<T> v) {
    cerr << "[  ";
    for (T i : v) {
      _print(i);
      cerr << "  ";
    }
    cerr << "]\n";
  }
  template <class T>
  void _print(multiset<T> v) {
    cerr << "[  ";
    for (T i : v) {
      _print(i);
      cerr << "  ";
    }
    cerr << "]\n";
  }
  template <class T, class V>
  void _print(map<T, V> v) {
    cerr << "[  ";
    for (auto i : v) {
      _print(i);
      cerr << "  ";
    }
    cerr << "]\n";
  }
  template <class T>
  void _print(T v[], long long n) {
    cerr << "[  ";
    for (long long i = 0; i < n; i++) {
      _print(v[i]);
      cerr << "  ";
    }
    cerr << "]\n";
  }
  struct BIT {
    vector<long long> values;
    long long N, LOGN;
    void init(long long n) {
      N = n;
      LOGN = ceil(log2(N));
      values.assign(N + 1, 0);
    }
    void update(long long idx, long long val) {
      while (idx <= N) {
        values[idx] += val;
        idx += idx & -idx;
      }
    }
    long long sum(long long idx) {
      long long sum = 0;
      while (idx > 0) {
        sum += values[idx];
        idx -= idx & -idx;
      }
      return sum;
    }
    long long rangeSum(long long i, long long j) {
      return sum(j) - (i == 1 ? 0 : sum(i - 1));
    }
  };
  void run() {
    long long n;
    cin >> n;
    long long a[n];
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
    }
    if (n == 1) {
      cout << "YES\n";
      return;
    }
    if (n == 2) {
      if (a[1] >= a[0])
        cout << "YES\n";
      else
        cout << "NO\n";
      return;
    }
    set<long long> s;
    for (long long i = 0; i < n; i++) s.insert(a[i]);
    if (s.size() != n) {
      cout << "YES\n";
      return;
    }
    BIT ds;
    ds.init(n + 10);
    long long ans = 0;
    for (long long i = n - 1; i >= 0; i--) {
      ans += ds.sum(a[i] - 1);
      ds.update(a[i], 1);
    }
    if (ans % 2 == 0)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0), cout.tie();
    long long t;
    cin >> t;
    for (long long tc = 1; tc <= t; tc++) {
      run();
    }
  }
  