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;
  mt19937 ran(time(0));
  long long mod = 1e9 + 7;
  long long gcd(long long a, long long b) {
    if (!b) return a;
    return gcd(b, a % b);
  }
  long long Bpow(long long a, long long b) {
    long long ans = 1;
    while (b) {
      if (b & 1) {
        ans *= a;
        ans %= mod;
      }
      a = a * a % mod;
      b >>= 1;
    }
    return ans;
  }
  map<pair<long long, long long>, long long> pos;
  void P(long long i, long long j, long long k,
         vector<pair<long long, long long>>& a) {
    pair<long long, long long> t = a[k];
    a[k] = a[j];
    a[j] = a[i];
    a[i] = t;
    pos[a[i]] = i;
    pos[a[j]] = j;
    pos[a[k]] = k;
  }
  int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    long long t;
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      vector<pair<long long, long long>> a;
      map<long long, long long> now;
      bool ff = false;
      for (long long i = 0; i < n; i++) {
        long long b;
        cin >> b;
        a.push_back({b, now[b]++});
        if (now[b] == 2) {
          ff = true;
        }
      }
      if (ff) {
        cout << "YES\n";
        continue;
      }
      vector<pair<long long, long long>> b = a;
      sort(b.begin(), b.end());
      for (long long i = 0; i < n; i++) {
        pos[a[i]] = i;
      }
      bool f = false;
      for (long long i = 0; i < n; i++) {
        if (a[i] == b[i]) continue;
        if (pos[b[i]] != i + 1) {
          P(i, i + 1, pos[b[i]], a);
        } else {
          if (i + 1 == n - 1) {
            if (a[i].first == a[i + 1].first) break;
            if (a[i + 1].second) {
              P(i, i + 1, pos[{a[i + 1].first, a[i + 1].second - 1}], a);
              break;
            }
            f = true;
            break;
          }
          P(i, i + 2, pos[b[i]], a);
        }
      }
      if (f)
        cout << "NO\n";
      else
        cout << "YES\n";
    }
  }
  