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;
  struct fenwickTree {
    vector<long long> bit;
    fenwickTree(long long n) : bit(n, 0) {}
    void update(long long idx, long long d) {
      for (; idx < (long long)bit.size(); idx = idx | (idx + 1)) bit[idx] += d;
    }
    long long sum(long long r) const {
      long long ret = 0;
      for (; r >= 0; r = (r & (r + 1)) - 1) ret += bit[r];
      return ret;
    }
    long long sum(long long l, long long r) const { return sum(r) - sum(l - 1); }
  };
  long long numIverserions(const vector<long long>& arr) {
    const long long n = arr.size();
    fenwickTree ft(n + 1);
    long long inversions = 0;
    for (long long val : arr) {
      if (val + 1 <= n) inversions += ft.sum(val + 1, n);
      ft.update(val, 1);
    }
    return inversions;
  }
  signed main() {
    cin.tie(0)->sync_with_stdio(false);
    long long t;
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      vector<long long> arr(n);
      vector<bool> cnts(n + 1, false);
      {
        bool isDup = false;
        for (long long& val : arr) {
          cin >> val;
          if (cnts[val]) {
            isDup = true;
          }
          cnts[val] = true;
        }
        if (isDup) {
          cout << "YES\n";
          continue;
        }
      }
      (numIverserions(arr) % 2 == 0 && cout << "YES\n") || cout << "NO\n";
    }
    return 0;
  }
  