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 <class T>
  using Graph = vector<vector<T>>;
  template <class T>
  using Maxheap = priority_queue<T>;
  template <class T>
  using Minheap = priority_queue<T, vector<T>, greater<T>>;
  template <typename T, typename T1>
  inline T max(T a, T1 b) {
    if (b > a) a = b;
    return a;
  }
  template <typename T, typename T1>
  inline T min(T a, T1 b) {
    if (b < a) a = b;
    return a;
  }
  inline int64_t ssum(int64_t n) { return (n * (n + 1)) / 2; };
  inline int64_t fastpow(int64_t a, int64_t b, int64_t m = (int64_t)(1e9 + 7)) {
    int64_t res = 1;
    a %= m;
    while (b > 0) {
      if (b & 1) res = (res * a) % m;
      a = (a * a) % m;
      b >>= 1;
    }
    return res;
  }
  void go() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
  }
  struct Tree {
    int64_t sz;
    vector<int64_t> bit;
    Tree(int64_t n) {
      sz = n;
      bit.assign(n, 0);
    }
    int64_t sum(int64_t i) {
      int64_t ans = 0;
      while (i >= 0) {
        ans += bit[i];
        i = (i & (i + 1)) - 1;
      }
      return ans;
    }
    void add(int64_t i, int64_t val) {
      while (i < sz) {
        bit[i] += val;
        i |= (i + 1);
      }
    }
  };
  signed main() {
    go();
    int64_t T = 1;
    cin >> T;
    for (int64_t tt = 1; tt <= T; tt++) {
      int64_t n;
      cin >> n;
      vector<int64_t> a(n);
      for (int64_t i = 0; i < n; i++) {
        cin >> a[i];
      }
      if (set<int64_t>(a.begin(), a.end()).size() < n) {
        cout << "YES\n";
      } else {
        Tree t(*max_element(a.begin(), a.end()) + 1);
        int64_t S = 0;
        for (int64_t i = n - 1; i >= 0; i--) {
          t.add(a[i], 1);
          S += t.sum(a[i] - 1);
        }
        cout << ((S & 1) ? "NO\n" : "YES\n");
      }
    }
    return 0;
  }
  