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 DIM = 5e5 + 7;
  const int INF = 7e18 + 7;
  const int mod = 998244353;
  const int base = 29;
  int n, t, k;
  int a[DIM];
  int b[DIM];
  int pos[DIM];
  set<int> s[DIM];
  void del(int type, int var) { s[type].erase(var); }
  void clr(int type, int var) {
    while (s[type].size() && (*s[type].begin()) <= var)
      s[type].erase(s[type].begin());
  }
  void add(int type, int var) { s[type].insert(var); }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin >> t;
    while (t--) {
      cin >> n;
      set<int> all;
      for (int i = 1; i <= n; i++) {
        cin >> a[i];
        all.insert(a[i]);
        b[i] = a[i];
        s[a[i]].insert(i);
      }
      if (all.size() != n) {
        cout << "YES\n";
        for (int i = 1; i <= n; i++) {
          s[b[i]].clear();
        }
        continue;
      }
      sort(b + 1, b + n + 1);
      int good = 1;
      int el1, el2, el3, k1, k2, k3;
      for (int i = 1; i <= n; i++) {
        if (b[i] == a[i]) {
          clr(a[i], i);
          continue;
        }
        if (i == n - 1) {
          good = 0;
          break;
        };
        el1 = a[i];
        k1 = i;
        el2 = b[i];
        k2 = (*s[el2].begin());
        el3 = a[i + 1];
        if (k2 == i + 1) el3 = a[i + 2];
        k3 = (*s[el3].rbegin());
        a[k1] = el2;
        a[k2] = el3;
        a[k3] = el1;
        del(el1, k1);
        del(el2, k2);
        del(el3, k3);
        add(el2, k1);
        add(el3, k2);
        add(el1, k3);
        clr(a[i], i);
      }
      for (int i = 1; i <= n; i++) {
        s[b[i]].clear();
      }
      if (good)
        cout << "YES\n";
      else
        cout << "NO\n";
    }
    return 0;
  }
  