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(long long t) { cerr << t; }
  void _print(string t) { cerr << t; }
  void _print(char t) { cerr << t; }
  void _print(long double t) { cerr << t; }
  void _print(double t) { cerr << t; }
  template <class T, class V>
  void _print(pair<T, V> p);
  template <class T>
  void _print(vector<T> v);
  template <class T>
  void _print(set<T> v);
  template <class T, class V>
  void _print(map<T, V> v);
  template <class T>
  void _print(multiset<T> v);
  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(vector<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(set<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(multiset<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T, class V>
  void _print(map<T, V> v) {
    cerr << "[ ";
    for (auto i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  long long mpow(long long base, long long exp);
  void ipgraph(long long m);
  void dfs(long long u, long long par);
  const long long mod = 1000000007;
  long long par[500000 + 5];
  long long rnk[500000 + 5];
  void make_set(long long n) {
    for (long long i = 1; i <= n; ++i) {
      par[i] = i;
      rnk[i] = 1;
    }
  }
  long long find_par(long long a) {
    if (par[a] == a) return a;
    return par[a] = find_par(par[a]);
  }
  void make_union(long long a, long long b) {
    long long p1 = find_par(a);
    long long p2 = find_par(b);
    if (p1 == p2) {
      return;
    }
    if (rnk[p1] > rnk[p2]) {
      par[p2] = p1;
      rnk[p1] += rnk[p2];
    } else {
      par[p1] = p2;
      rnk[p2] += rnk[p1];
    }
  }
  void sol() {
    long long n, i, j, k, m, q;
    bool pos = true;
    cin >> n;
    make_set(n);
    std::vector<long long> a(n, 0);
    std::vector<pair<long long, long long> > v, v_index;
    unordered_map<long long, long long> make_pair;
    unordered_set<long long> st;
    for (i = 0; i < n; i++) {
      cin >> a[i];
      v.push_back({a[i], i + 1});
      st.insert(a[i]);
    }
    sort(v.begin(), v.end());
    std::vector<long long> b = a;
    sort(b.begin(), b.end());
    if (a == b) {
      cout << "YES" << '\n';
      return;
    }
    for (long long i = 0; i < n; i++) {
      make_union(v[i].second, i + 1);
    }
    std::vector<bool> done(n + 5, false);
    long long even = 0;
    for (long long i = 1; i <= n; i++) {
      long long temp = rnk[find_par(i)];
      if (find_par(i) == i) {
        if (temp % 2 == 0) {
          even++;
          pos = false;
        }
      }
    }
    if (st.size() == a.size() && even % 2 == 1) {
      cout << "NO" << '\n';
      return;
    }
    cout << "YES" << '\n';
  }
  int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long t;
    cin >> t;
    while (t--) sol();
    return 0;
  }
  long long mpow(long long base, long long exp) {
    base %= mod;
    long long result = 1;
    while (exp > 0) {
      if (exp & 1) result = (result * base) % mod;
      base = (base * base) % mod;
      exp >>= 1;
    }
    return result;
  }
  