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,no-stack-protector,tune=native")
  #pragma gcc optimize("sse,sse2,sse3,sse4,ssse3")
  #pragma gcc optimize("O2")
  #pragma gcc optimize("abm,mmx,avx,avx2,unroll-loops,fast-math,section-anchors")
  using namespace std;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  void solve() {
    int n;
    cin >> n;
    vector<int> a(n), in(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      a[i]--;
    }
    if (n < 3) {
      cout << ((n == 1 || a[0] <= a[1]) ? "YES\n" : "NO\n");
      return;
    }
    vector<int> p(n, 0), val(n, -1);
    set<int> wrong;
    for (int i = 0; i < n; i++) {
      p[a[i]] = 1;
      val[a[i]] = i;
      if (a[i] != i) wrong.insert(i);
    }
    int sum = accumulate(p.begin(), p.end(), 0);
    ;
    if (sum != n) {
      cout << ((1) ? "YES\n" : "NO\n");
      return;
    }
    for (int val2_ = 0; val2_ < n; val2_++) {
      int val2 = val2_;
      if (wrong.find(val2) == wrong.end()) continue;
      int pos2 = val[val2];
      int pos1 = val2;
      int val1 = a[pos1];
      auto it = wrong.begin();
      while (it != wrong.end() && (*it == pos1 || *it == pos2)) {
        it++;
      }
      if (it == wrong.end()) {
        cout << ((0) ? "YES\n" : "NO\n");
        return;
      }
      int pos3 = *it, val3 = a[pos3];
      if ((pos2 > pos1 && pos2 > pos3 && pos1 < pos3) ||
          (pos2 < pos1 && pos2 < pos3 && pos1 < pos3)) {
        vector<pair<int, int> > cur = {{pos1, val1}, {pos2, val2}, {pos3, val3}};
        sort(cur.begin(), cur.end());
        tie(pos1, val1) = cur[0];
        tie(pos2, val2) = cur[1];
        tie(pos3, val3) = cur[2];
        a[pos2] = val1;
        a[pos3] = val2;
        a[pos1] = val3;
        val[val1] = pos2;
        val[val2] = pos3;
        val[val3] = pos1;
        if (val1 == pos2) wrong.erase(pos2);
        if (val2 == pos3) wrong.erase(pos3);
        if (val3 == pos1) wrong.erase(pos1);
      } else {
        vector<pair<int, int> > cur = {{pos1, val1}, {pos2, val2}, {pos3, val3}};
        sort(cur.begin(), cur.end());
        tie(pos1, val1) = cur[0];
        tie(pos2, val2) = cur[1];
        tie(pos3, val3) = cur[2];
        a[pos1] = val2;
        a[pos2] = val3;
        a[pos3] = val1;
        val[val2] = pos1;
        val[val3] = pos2;
        val[val1] = pos3;
        if (val2 == pos1) wrong.erase(pos1);
        if (val3 == pos2) wrong.erase(pos2);
        if (val1 == pos3) wrong.erase(pos3);
      }
    }
    cout << ((1) ? "YES\n" : "NO\n");
    return;
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    cin >> t;
    while (t--) solve();
  }
  