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;
  const int N = 5e5 + 5;
  struct FT {
    vector<long long> s;
    FT(int n) : s(n) {}
    void upd(int pos, long long dif) {
      for (; pos < (int)(s).size(); pos |= pos + 1) s[pos] += dif;
    }
    long long qry(int pos) {
      long long res = 0;
      for (; pos > 0; pos &= pos - 1) res += s[pos - 1];
      return res;
    }
    int lower_bound(long long sum) {
      if (sum <= 0) return -1;
      int pos = 0;
      for (int pw = 1 << 25; pw; pw >>= 1) {
        if (pos + pw <= (int)(s).size() && s[pos + pw - 1] < sum)
          pos += pw, sum -= s[pos - 1];
      }
      return pos;
    }
  };
  int n, a[N];
  bool vis[N];
  void solve() {
    cin >> n;
    fill(vis, vis + n, 0);
    bool ok = false;
    for (int i = 0; i < (n); ++i) {
      cin >> a[i];
      --a[i];
      if (vis[a[i]]) {
        ok = true;
      } else {
        vis[a[i]] = true;
      }
    }
    FT ft(n);
    long long cnt = 0;
    for (int i = 0; i < (n); ++i) {
      cnt += ft.qry(n) - ft.qry(a[i] + 1);
      ft.upd(a[i], 1);
    }
    cout << (cnt % 2 == 0 || ok ? "YES\n" : "NO\n");
  }
  int main() {
    cin.tie(0)->sync_with_stdio(0);
    cin.exceptions(cin.failbit);
    int t;
    cin >> t;
    while (t--) {
      solve();
    }
  }
  