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>
  #pragma GCC optimize "trapv"
  using namespace std;
  template <typename T1, typename T2>
  istream &operator>>(istream &istream, pair<T1, T2> &p) {
    return (istream >> p.first >> p.second);
  }
  template <typename T>
  istream &operator>>(istream &istream, vector<T> &v) {
    for (auto &it : v) cin >> it;
    return istream;
  }
  template <typename T1, typename T2>
  ostream &operator<<(ostream &ostream, const pair<T1, T2> &p) {
    return (ostream << p.first << " " << p.second);
  }
  template <typename T>
  ostream &operator<<(ostream &ostream, const vector<T> &c) {
    for (auto &it : c) cout << it << " ";
    return ostream;
  }
  vector<pair<int, int> > dir = {{0, 1}, {1, 0}, {-1, 0}, {0, -1}};
  struct DSU {
    vector<int> par;
    vector<int> sze;
    int connected;
    void ini(int n) {
      par.resize(n + 1);
      sze.assign(n + 1, 1);
      iota((par).begin(), (par).end(), 0);
      connected = n;
    }
    int get(int x) { return (par[x] = (par[x] == x ? x : get(par[x]))); }
    int getSize(int u) { return sze[get(u)]; }
    void un(int x, int y) {
      x = get(x);
      y = get(y);
      if (x == y) {
        return;
      }
      sze[y] += sze[x];
      sze[x] = 0;
      par[x] = y;
      connected--;
    }
  };
  void solve() {
    int n;
    cin >> n;
    vector<int> v(n);
    cin >> v;
    vector<int> vc = v;
    sort((vc).begin(), (vc).end());
    map<int, int> m;
    for (int i = 0; 0 < n ? i < n : i > n; 0 < n ? i += 1 : i -= 1) m[v[i]]++;
    if (m.size() < n) {
      cout << "YES\n";
      return;
    }
    DSU dsu;
    dsu.ini(n + 1);
    for (int i = 0; 0 < n ? i < n : i > n; 0 < n ? i += 1 : i -= 1)
      dsu.un(v[i], i + 1);
    int s = 0;
    map<int, int> f;
    for (int i = 1; 1 < n + 1 ? i < n + 1 : i > n + 1;
         1 < n + 1 ? i += 1 : i -= 1) {
      if (f[dsu.get(i)] == 0 && dsu.getSize(i) % 2 == 0) {
        s++;
        f[dsu.get(i)] = 1;
      }
    }
    if (s % 2)
      cout << "NO";
    else
      cout << "YES";
    cout << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
    cout << fixed << setprecision(0);
    int t = 1;
    cin >> t;
    for (int i = 1; i <= t; i++) {
      solve();
    };
    return 0;
  }
  