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 int mod = 1000000009;
  vector<int> seg;
  int seg_q(int num, int le, int ri, int x, int y) {
    if (y < le || ri < x)
      return 0;
    else if (x <= le && ri <= y) {
      return seg[num];
    }
    int mid = (le + ri) / 2;
    return seg_q(num * 2, le, mid, x, y) + seg_q(num * 2 + 1, mid + 1, ri, x, y);
  }
  int seg_u(int num, int le, int ri, int x) {
    if (x < le || ri < x)
      return seg[num];
    else if (x == le && le == ri)
      return ++seg[num];
    int mid = (le + ri) / 2;
    return seg[num] =
               seg_u(num * 2, le, mid, x) + seg_u(num * 2 + 1, mid + 1, ri, x);
  }
  void init() {}
  void solve() {
    int n;
    cin >> n;
    seg = vector<int>(n * 4 + 4);
    long long cnt = 0;
    bool ch = false;
    for (int i = 0; i < (int)(n); ++i) {
      int x;
      cin >> x;
      cnt += seg_q(1, 1, n, x + 1, n);
      if (!ch && seg_q(1, 1, n, x, x) > 0) ch = true;
      seg_u(1, 1, n, x);
    }
    cout << (cnt % 2 && !ch ? "NO" : "YES") << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    ;
    init();
    int test_cases = 1;
    cin >> test_cases;
    while (test_cases--) {
      solve();
    }
    return 0;
  }
  