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>
  #pragma GCC optimize("Ofast")
  #pragma GCC target("avx,avx2,fma")
  #pragma GCC optimization("unroll-loops")
  using namespace std;
  long long merge(long long arr[], long long l, long long mid, long long r) {
    long long ans = 0;
    long long n1 = mid - l + 1;
    long long n2 = r - mid;
    long long a[n1];
    long long b[n2];
    for (long long i = 0; i < n1; i++) {
      a[i] = arr[l + i];
    }
    for (long long i = 0; i < n2; i++) {
      b[i] = arr[mid + 1 + i];
    }
    long long i = 0;
    long long j = 0;
    long long k = l;
    while (i < n1 && j < n2) {
      if (a[i] <= b[j]) {
        arr[k] = a[i];
        k++;
        i++;
      } else {
        ans += n1 - i;
        arr[k] = b[j];
        k++;
        j++;
      }
    }
    while (i < n1) {
      arr[k] = a[i];
      k++;
      i++;
    }
    while (j < n2) {
      arr[k] = b[j];
      k++;
      j++;
    }
    return ans;
  }
  long long inversion_count(long long arr[], long long l, long long r) {
    long long ans = 0;
    if (l < r) {
      long long mid = (l + r) / 2;
      ans += inversion_count(arr, l, mid);
      ans += inversion_count(arr, mid + 1, r);
      ans += merge(arr, l, mid, r);
    }
    return ans;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    long long t;
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      long long v[n];
      for (long long i = 0; i < n; i++) cin >> v[i];
      bool ans = false;
      long long inv = inversion_count(v, 0, n - 1);
      if (inv % 2 == 0)
        ans = true;
      else {
        sort(v, v + n);
        for (long long i = 0; i < n - 1; i++)
          if (v[i] == v[i + 1]) ans = true;
      }
      cout << (ans ? "YES" : "NO") << '\n';
    }
    return 0;
  }
  