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;
  const long long MOD = 1e9 + 7;
  const long long INF = 1e9;
  const long double EPS = 1e-9;
  const long long N = 1e6 + 2;
  long long mod(long long x, long long y = MOD) { return ((x % y) + y) % y; }
  long long addm(long long x) { return x; }
  template <typename... Ts>
  long long addm(long long x, Ts... ys) {
    return mod(x + addm(ys...));
  }
  long long mulm(long long x) { return x; }
  template <typename... Ts>
  long long mulm(long long x, Ts... ys) {
    return mod(x * mulm(ys...));
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--) {
      int n;
      cin >> n;
      pair<long long, long long> arr[n];
      vector<long long> loop;
      int visited[n];
      memset(visited, 0, sizeof visited);
      ;
      for (long long i = (0); i < (n); i++) {
        int a;
        cin >> a;
        arr[i] = pair<long long, long long>(a, i);
      }
      sort(arr, arr + n);
      bool flag = false;
      for (long long i = (1); i < (n); i++) {
        if (arr[i].first == arr[i - 1].first) flag = true;
      }
      if (flag) {
        cout << "YES\n";
        continue;
      }
      int count = 0;
      int oddcount = 0;
      for (long long i = (0); i < (n); i++) {
        if (visited[i]) continue;
        int u = i;
        int cnt = 0;
        while (!visited[u]) {
          cnt++;
          visited[u] = 1;
          u = arr[u].second;
        }
        if (cnt % 2 == 0)
          count++;
        else if (cnt != 1)
          oddcount++;
      }
      if (count % 2 == 1)
        cout << "NO\n";
      else
        cout << "YES\n";
    }
    return 0;
  }
  