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 BIT {
    vector<T> v;
    int n;
    BIT(int n) : n(n), v(n) {}
    void add(int x, T val) {
      for (int i = x + 1; i <= n; i += i & -i) v[i - 1] += val;
    }
    T qry(int x) {
      T res = 0;
      for (int i = min(x + 1, n); i > 0; i -= i & -i) res += v[i - 1];
      return res;
    }
  };
  void solve() {
    int n;
    cin >> n;
    vector<int> v(n);
    for (auto &d : v) cin >> d;
    vector<int> vv = v;
    sort(begin(vv), end(vv));
    vv.resize(unique(begin(vv), end(vv)) - begin(vv));
    BIT<int> fen(vv.size());
    long long ans = 0;
    for (int i = 0; i < v.size(); i++) {
      int pos = lower_bound(begin(vv), end(vv), v[i]) - begin(vv);
      ans += i - fen.qry(pos);
      fen.add(pos, 1);
    }
    cout << (ans % 2 == 0 || set<int>(begin(v), end(v)).size() != n ? "Yes"
                                                                    : "No")
         << "\n";
  }
  int main() {
    int t;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  