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;
  template <class T>
  bool uin(T &a, T b) {
    return a > b ? (a = b, true) : false;
  }
  template <class T>
  bool uax(T &a, T b) {
    return a < b ? (a = b, true) : false;
  }
  mt19937 rng(chrono::system_clock::now().time_since_epoch().count());
  const int maxN = 5e5 + 10;
  template <class T, class FApply>
  struct FenwickTree {
    T const identity;
    std::vector<T> f;
    FApply apply;
    FenwickTree(FenwickTree const &other) = default;
    FenwickTree(int n, T const identity, FApply const &apply)
        : identity(identity), f(n + 1, identity), apply(apply) {}
    void init(int n) { f.assign(n + 1, identity); }
    void update(int x, T d) {
      for (; x < (int)f.size(); x += x & -x) {
        apply(f[x], d);
      }
    }
    T prefix(int x) {
      T ans(identity);
      for (; x > 0; x &= x - 1) {
        apply(ans, f[x]);
      }
      return ans;
    }
  };
  template <class T, class FApply>
  FenwickTree<T, FApply> makeFenwickTree(int const &n, T const &identity,
                                         FApply const &apply) {
    return FenwickTree<T, FApply>(n, identity, apply);
  }
  int n, a[maxN];
  void solve() {
    cin >> n;
    for (int i = 1; i <= (int)(n); ++i) {
      cin >> a[i];
    }
    map<int, int> mp;
    for (int i = 1; i <= (int)(n); ++i) {
      mp[a[i]]++;
    }
    for (auto &p : mp) {
      if (p.second >= 2) {
        cout << "YES\n";
        return;
      }
    }
    auto f = makeFenwickTree<int>(n, 0, [](int &x, int y) { x += y; });
    int h = 0;
    for (int i = n; i; i--) {
      h ^= (f.prefix(a[i]) & 1);
      f.update(a[i], 1);
    }
    cout << (h ? "NO" : "YES") << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int T = 1;
    cin >> T;
    while (T--) {
      solve();
    }
    return 0;
  }
  