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;
  int arr[500005], brr[500005];
  pair<int, int> parr[500005];
  int ord[500005];
  int dfs(int idx) {
    if (ord[idx] != 0) return 0;
    int ret = 1;
    vector<int> same_ord;
    int nex = idx;
    while (brr[nex] != idx) {
      nex = brr[nex];
      same_ord.push_back(nex);
      ret += 1;
    }
    for (int i = 0; i < same_ord.size(); i++) {
      ord[same_ord[i]] = ret;
    }
    ord[idx] = ret;
    return ret;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T;
    cin >> T;
    while (T--) {
      int n;
      cin >> n;
      for (int i = 0; i < n; i++) {
        cin >> parr[i].first;
        parr[i].second = i;
        ord[i + 1] = 0;
      }
      sort(parr, parr + n);
      int ev = 0;
      for (int i = 0; i < n - 1; i++) {
        if (parr[i].first == parr[i + 1].first) {
          goto YES;
        }
      }
      for (int i = 0; i < n; i++) {
        arr[i + 1] = parr[i].second + 1;
      }
      for (int i = 1; i <= n; i++) {
        brr[arr[i]] = i;
      }
      for (int i = 1; i < n + 1; i++) {
        int x = dfs(i);
        if (x == 0)
          continue;
        else {
          if (x % 2 == 0) ev += 1;
        }
      }
    YES:
      if (ev % 2 == 0)
        cout << "YES" << '\n';
      else
        cout << "NO" << '\n';
    }
  }
  