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;
  #pragma warning(disable : 4996)
  const long long N = 2e5 + 5;
  const long long mod = 1e9 + 7;
  const long long inf = 1e18;
  long long _mergeSort(vector<long long>& arr, vector<long long>& temp,
                       long long left, long long right);
  long long merge(vector<long long>& arr, vector<long long>& temp, long long left,
                  long long mid, long long right);
  long long mergeSort(vector<long long> arr) {
    vector<long long> temp(((long long)arr.size()));
    return _mergeSort(arr, temp, 0, ((long long)arr.size()) - 1);
  }
  long long _mergeSort(vector<long long>& arr, vector<long long>& temp,
                       long long left, long long right) {
    long long mid, inv_count = 0;
    if (right > left) {
      mid = (right + left) / 2;
      inv_count += _mergeSort(arr, temp, left, mid);
      inv_count += _mergeSort(arr, temp, mid + 1, right);
      inv_count += merge(arr, temp, left, mid + 1, right);
    }
    return inv_count;
  }
  long long merge(vector<long long>& arr, vector<long long>& temp, long long left,
                  long long mid, long long right) {
    long long i, j, k;
    long long inv_count = 0;
    i = left;
    j = mid;
    k = left;
    while ((i <= mid - 1) && (j <= right)) {
      if (arr[i] <= arr[j]) {
        temp[k++] = arr[i++];
      } else {
        temp[k++] = arr[j++];
        inv_count = inv_count + (mid - i);
      }
    }
    while (i <= mid - 1) temp[k++] = arr[i++];
    while (j <= right) temp[k++] = arr[j++];
    for (i = left; i <= right; i++) arr[i] = temp[i];
    return inv_count;
  }
  void solve() {
    long long n;
    cin >> n;
    vector<long long> a(n), b(n);
    map<long long, long long> pos;
    map<long long, long long> f;
    long long flag = 0;
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
      pos[a[i]] = i;
      b[i] = a[i];
      f[a[i]]++;
      if (f[a[i]] == 2) {
        flag = 1;
      }
    }
    long long x = mergeSort(a);
    if (x % 2 == 0) {
      cout << "YES";
    } else if (flag == 1) {
      cout << "YES";
    } else if (x % 2 == 1) {
      cout << "NO";
    }
    return;
    sort(b.begin(), b.end());
    for (long long i = 0; i < n - 3; i++) {
      if (i < n - 3) {
        if (a[i] == b[i]) {
          continue;
        } else {
          long long p = pos[b[i]];
          if (p == i + 1) {
            long long temp1 = a[i];
            long long temp2 = a[i + 1];
            long long temp3 = a[i + 2];
            pos[a[i]] = i + 2;
            pos[a[i + 1]] = i;
            pos[a[i + 2]] = i + 1;
            a[i] = temp2;
            a[i + 1] = temp3;
            a[i + 2] = temp1;
          } else {
            long long temp1 = a[i];
            long long temp2 = a[i + 1];
            pos[a[i]] = i + 1;
            pos[b[i]] = i;
            pos[a[i + 1]] = p;
            a[i] = b[i];
            a[i + 1] = temp1;
            a[p] = temp2;
          }
        }
      }
    }
    long long r = n % 3;
    if (r == 1) {
      cout << "YES";
    } else if (r == 2) {
      if (a[n - 2] <= a[n - 1]) {
        cout << "YES";
      } else {
        cout << "NO";
      }
    } else {
      long long n1 = a[n - 3];
      long long n2 = a[n - 2];
      long long n3 = a[n - 1];
      if (n1 <= n2 && n2 <= n3) {
        cout << "YES";
      } else if (n2 >= n1 && n2 >= n3 && n2 >= n1) {
        cout << "YES";
      } else if (n1 >= n2 && n1 >= n3 && n3 >= n2) {
        cout << "YES";
      } else {
        cout << "NO";
      }
    }
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(0);
    long long t = 0;
    long long test_cases = 1;
    cin >> test_cases;
    while (t++ < test_cases) {
      solve();
      cout << "\n";
    }
    cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs"
         << "\n";
    return 0;
  }
  