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 long long MOD = 998244353;
  const long long INF = 1e18;
  const long long N = 500005;
  int t[4 * N];
  void update(int v, int l, int r, int pos) {
    if (l == r) {
      t[v] = 1 - t[v];
      return;
    }
    int m = (l + r) / 2;
    if (pos <= m) {
      update(2 * v, l, m, pos);
    } else {
      update(2 * v + 1, m + 1, r, pos);
    }
    t[v] = t[2 * v] + t[2 * v + 1];
  }
  int get(int v, int l, int r, int s, int e) {
    if (s > e) {
      return 0;
    }
    if (l == s and r == e) {
      return t[v];
    }
    int m = (l + r) / 2;
    return get(2 * v, l, m, s, min(m, e)) +
           get(2 * v + 1, m + 1, r, max(m + 1, s), e);
  }
  bool b[N];
  long long x[N];
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    ;
    int tt;
    cin >> tt;
    while (tt--) {
      long long n;
      cin >> n;
      bool check = false;
      long long ans = 0;
      for (long long i = (1); i < (n + 1); i++) {
        b[i] = 0;
      }
      for (long long i = (1); i < (n + 1); i++) {
        cin >> x[i];
        if (b[x[i]]) {
          check = true;
        }
        b[x[i]] = 1;
      }
      if (check) {
        cout << "YES" << endl;
        continue;
      }
      update(1, 1, n, x[1]);
      for (long long i = (2); i < (n + 1); i++) {
        ans += get(1, 1, n, x[i] + 1, n);
        update(1, 1, n, x[i]);
      }
      for (long long i = (1); i < (n + 1); i++) {
        update(1, 1, n, x[i]);
      }
      cout << "NO\0YES" + 3 * (ans % 2 == 0) << endl;
    }
    return 0;
  }
  