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 long long MOD = 1e9 + 7;
  const long long N = 2e5 + 10;
  long long ans = 0;
  void mergeSort(vector<long long> &v, int st, int end) {
    if (st == end) {
      return;
    }
    int mid = (st + end) / 2;
    mergeSort(v, st, mid);
    mergeSort(v, mid + 1, end);
    vector<int> a, b;
    for (int i = st; i <= mid; i++) {
      a.push_back(v[i]);
    }
    for (int i = mid + 1; i <= end; i++) {
      b.push_back(v[i]);
    }
    int i = 0, j = 0, curr = st;
    int aa = a.size(), bb = b.size();
    while (i < aa or j < bb) {
      if (i < aa and j < bb and a[i] <= b[j]) {
        v[curr++] = a[i++];
      } else if (i < aa and j < bb and a[i] > b[j]) {
        v[curr++] = b[j++];
        ans += aa - i;
      } else if (i < aa) {
        v[curr++] = a[i++];
      } else {
        v[curr++] = b[j++];
      }
    }
  }
  long long countInversions(vector<long long> &v, int n) {
    mergeSort(v, 0, n - 1);
    return ans;
  }
  void test_case() {
    long long n;
    cin >> n;
    vector<long long> v(n);
    for (int i = 0; i < n; i++) {
      cin >> v[i];
    }
    ans = 0;
    countInversions(v, n);
    set<long long> st(v.begin(), v.end());
    if (st.size() < n or ans % 2 == 0) {
      cout << "YES\n";
    } else {
      cout << "NO\n";
    }
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int tc = 1;
    cin >> tc;
    while (tc--) {
      test_case();
    }
    return 0;
  }
  