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;
  void solve() {
    int n;
    cin >> n;
    vector<int> a(n);
    vector<pair<int, int>> b(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      b[i].first = a[i];
      b[i].second = i;
    }
    sort(b.begin(), b.end());
    for (int i = 1; i < n; i++) {
      if (b[i].first == b[i - 1].first) {
        cout << "YES\n";
        return;
      }
    }
    vector<int> pos(n);
    for (int i = 0; i < n; i++) {
      a[b[i].second] = i;
    }
    for (int i = 0; i < n; i++) {
      pos[a[i]] = i;
    }
    for (int i = 0; i < n - 2; i++) {
      if (a[i] == i)
        continue;
      else if (a[i] != n - 1 && a[i] != n - 2) {
        pos[a[i]] = pos[n - 1];
        pos[n - 1] = pos[i];
        a[pos[n - 1]] = n - 1;
        a[pos[a[i]]] = a[i];
      } else if (a[i] == n - 1) {
        pos[n - 1] = pos[n - 2];
        pos[n - 2] = pos[i];
        a[pos[n - 1]] = n - 1;
        a[pos[n - 2]] = n - 2;
      } else if (a[i] == n - 2) {
        pos[n - 2] = pos[n - 1];
        pos[n - 1] = pos[i];
        a[pos[n - 1]] = n - 1;
        a[pos[n - 2]] = n - 2;
      }
    }
    if (pos[n - 1] == n - 1) {
      cout << "YES\n";
    } else
      cout << "NO\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    while (t--) solve();
  }
  