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 <typename T>
  struct binary_indexed_tree {
    int N;
    vector<T> BIT;
    binary_indexed_tree(int N) : N(N), BIT(N + 1, 0) {}
    void add(int i, T x) {
      i++;
      while (i <= N) {
        BIT[i] += x;
        i += i & -i;
      }
    }
    T sum(int i) {
      T ans = 0;
      while (i > 0) {
        ans += BIT[i];
        i -= i & -i;
      }
      return ans;
    }
    T sum(int L, int R) { return sum(R) - sum(L); }
  };
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int t;
    cin >> t;
    for (int i = 0; i < t; i++) {
      int n;
      cin >> n;
      vector<int> a(n);
      for (int j = 0; j < n; j++) {
        cin >> a[j];
        a[j]--;
      }
      binary_indexed_tree<int> BIT(n);
      long long S = 0;
      for (int j = 0; j < n; j++) {
        S += BIT.sum(a[j] + 1, n);
        BIT.add(a[j], 1);
      }
      if (S % 2 == 0) {
        cout << "YES"
             << "\n";
      } else {
        sort(a.begin(), a.end());
        bool ok = false;
        for (int j = 0; j < n - 1; j++) {
          if (a[j] == a[j + 1]) {
            ok = true;
          }
        }
        if (ok) {
          cout << "YES"
               << "\n";
        } else {
          cout << "NO"
               << "\n";
        }
      }
    }
  }
  