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 M = 1e9 + 7;
  const int N = 2e5 + 5;
  long long power(long long x, long long y, long long p) {
    long long res = 1;
    x = x % p;
    while (y > 0) {
      if (y & 1) res = (res * x) % p;
      y = y >> 1;
      x = (x * x) % p;
    }
    return res;
  }
  long long mi(long long x) { return power(x, M - 2, M); }
  int nCr(int n, int r) {
    int mod = 1000000007;
    int C[r + 1];
    memset(C, 0, sizeof(C));
    C[0] = 1;
    for (int i = 1; i <= n; i++) {
      for (int j = min(i, r); j > 0; j--) {
        C[j] = (C[j] + C[j - 1]) % mod;
      }
    }
    return C[r] % mod;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long T;
    cin >> T;
    while (T--) {
      long long n;
      cin >> n;
      vector<long long> v(n);
      long long fg = 0;
      map<long long, long long> mp;
      vector<long long> v2;
      for (long long i = 0; i < n; i++) {
        cin >> v[i];
        if (mp[v[i]] != 0) {
          fg = 1;
        }
        mp[v[i]]++;
        v2.push_back(v[i]);
      }
      if (fg) {
        cout << "YES" << endl;
      } else {
        sort(v2.begin(), v2.end());
        map<long long, long long> m;
        for (long long i = 0; i < n; i++) {
          m[v2[i]] = i;
        }
        vector<long long> vis(n, 0);
        long long cn1 = 0, cn2 = 0;
        for (long long i = 0; i < n; i++) {
          if (m[v[i]] == i) {
            continue;
          }
          if (!vis[i]) {
            vis[i] = 1;
            long long pres = i;
            long long st = m[v[i]];
            long long cn = 1;
            while (st != pres) {
              vis[st] = 1;
              cn++;
              st = m[v[st]];
            }
            if (cn % 2 != 0) {
              cn1++;
            } else {
              cn2++;
            }
          }
        }
        if (cn2 % 2 == 0) {
          cout << "YES" << endl;
        } else {
          cout << "NO" << endl;
        }
      }
    }
    return 0;
  }
  