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;
  long long IOS = []() {
    ios::sync_with_stdio(0);
    std::cin.tie(nullptr);
    std::cout.tie(nullptr);
    return 0;
  }();
  template <class T>
  void _W(const T &x) {
    cout << x;
  }
  template <class T>
  void _W(T &x) {
    cout << x;
  }
  template <class T, class U>
  void _W(const pair<T, U> &x) {
    _W(x.first);
    putchar(' ');
    _W(x.second);
  }
  template <class T>
  void _W(const vector<T> &x) {
    for (auto i = x.begin(); i != x.end(); _W(*i++)) cout << ' ';
  }
  void W() {}
  template <class T, class... U>
  void W(const T &head, const U &...tail) {
    _W(head);
    cout << ", ";
    W(tail...);
  }
  using pii = pair<long long, long long>;
  const long long N = 2e6 + 10, mod = 998244353;
  inline long long quick(long long a, long long b) {
    a %= mod;
    long long res = 1;
    while (b) {
      if (b & 1) res = res * a % mod;
      a = a * a % mod;
      b >>= 1;
    }
    return res;
  }
  namespace Bit {
  constexpr long long MAXN = 1e6 + 5;
  long long n;
  long long t1[MAXN], t2[MAXN];
  void _add(long long l, long long c1) {
    long long c2 = c1 * l;
    for (long long i = l; i <= n; i += i & -i) t1[i] += c1, t2[i] += c2;
  }
  long long _qu1(long long i) {
    long long res = 0;
    for (; i > 0; i -= i & -i) res += t1[i];
    return res;
  }
  long long _qu2(long long i) {
    long long res = 0;
    for (; i > 0; i -= i & -i) res += t2[i];
    return res;
  }
  void init(long long _n) {
    n = _n;
    fill(t1, t1 + n + 1, 0);
    fill(t2, t2 + n + 1, 0);
  }
  void add(long long l, long long r, long long c) { _add(l, c), _add(r + 1, -c); }
  long long sum(long long l, long long r) {
    return (r + 1LL) * _qu1(r) - l * _qu1(l) - (_qu2(r) - _qu2(l));
  }
  long long query(long long x) { return _qu1(x); }
  }  // namespace Bit
  void solve() {
    long long n;
    cin >> n;
    Bit::init(n + 10);
    set<long long> s;
    long long cnt = 0;
    for (long long i = 1; i <= n; i++) {
      long long x;
      cin >> x;
      s.insert(x);
      cnt += Bit::sum(x + 1, n);
      Bit::add(x, x, 1);
    }
    if (s.size() != n || cnt % 2 == 0)
      cout << "YES" << '\n';
    else
      cout << "NO" << '\n';
  }
  signed main() {
    long long tt = 1;
    cin >> tt;
    while (tt--) {
      solve();
    }
  }
  