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 maxn = 500000 + 50;
  int T, n, h[maxn], f[maxn], tot, ti[maxn];
  int ans, sum;
  int main() {
    scanf("%d", &T);
    for (int oo = 1; oo <= T; ++oo) {
      scanf("%d", &n);
      tot = 0;
      vector<bool> used(n);
      vector<set<int>> gi(n), ed(n);
      vector<int> seet;
      ans = 0, sum = 0;
      for (int i = 0; i < n; i++) f[i] = 0;
      for (int i = 0; i < n; i++) {
        scanf("%d", &h[i]);
        tot = max(tot, ++f[--h[i]]);
      }
      for (int i = 0; i < n; ++i) ti[i] = h[i];
      sort(h, h + n);
      for (int i = 0; i < n; i++)
        if (h[i] != ti[i])
          gi[ti[i]].insert(i), seet.push_back(i), ed[h[i]].insert(i);
      function<void(int)> dfs = [&](int u) {
        ed[h[u]].erase(u);
        sum++;
        gi[ti[u]].erase(u);
        if (!ed[ti[u]].empty()) dfs(*ed[ti[u]].begin());
      };
      for (int i : seet)
        if (gi[ti[i]].count(i)) {
          sum = 0;
          dfs(i);
          if (sum % 2 == 0) ++ans;
        }
      bool p = 1;
      if (tot > 1 || ans % 2 == 0) p = 0;
      if (!p)
        printf("YES\n");
      else
        printf("NO\n");
    }
    return 0;
  }
  