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;
  int T, n, b[500005], a[500005], cnt;
  map<int, int> tu, pos;
  bool fl, vis[500005];
  void dfs(int x) {
    vis[x] = 1, cnt++;
    if (!vis[a[x]]) dfs(a[x]);
  }
  int main() {
    cin >> T;
    while (T--) {
      cin >> n;
      tu.clear(), fl = 0, cnt = 0;
      for (int i = 1; i <= n; i++) cin >> a[i], tu[a[i]]++;
      for (int i = 1; i <= n; i++)
        if (tu[a[i]] > 1) fl = 1;
      if (n == 1 || fl) {
        puts("YES");
        continue;
      }
      for (int i = 1; i <= n; i++) b[i] = a[i];
      sort(b + 1, b + 1 + n);
      for (int i = 1; i <= n; i++) pos[b[i]] = i;
      for (int i = 1; i <= n; i++) a[i] = pos[a[i]], vis[i] = 0;
      for (int i = 1; i <= n; i++)
        if (!vis[i]) dfs(i), cnt--;
      puts(((cnt)&1) ? "NO" : "YES");
    }
  }
  