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;
  void swap(int &x, int &y) {
    int temp = x;
    x = y;
    y = temp;
  }
  int XOR(int n) {
    if (n % 4 == 0) return n;
    if (n % 4 == 1) return 1;
    if (n % 4 == 2) return n + 1;
    return 0;
  }
  int isPrime(int N) {
    if (N < 2 || (!(N & 1) && N != 2)) return 0;
    for (int i = 3; i * i <= N; i += 2) {
      if (!(N % i)) return 0;
    }
    return 1;
  }
  void fillPrimes(vector<int> &prime, int high) {
    bool ck[high + 1];
    memset(ck, true, sizeof(ck));
    ck[1] = false;
    ck[0] = false;
    for (int i = 2; (i * i) <= high; i++) {
      if (ck[i] == true) {
        for (int j = i * i; j <= high; j = j + i) {
          ck[j] = false;
        }
      }
    }
    for (int i = 2; i * i <= high; i++) {
      if (ck[i] == true) {
        prime.push_back(i);
      }
    }
  }
  vector<bool> segmentedSieve(int low, int high) {
    vector<bool> prime(high - low + 1, true);
    vector<int> chprime;
    fillPrimes(chprime, high);
    for (int i : chprime) {
      int lower = (low / i);
      if (lower <= 1) {
        lower = i + i;
      } else if (low % i) {
        lower = (lower * i) + i;
      } else {
        lower = (lower * i);
      }
      for (int j = lower; j <= high; j = j + i) {
        prime[j - low] = false;
      }
    }
    return prime;
  }
  const double eps = 1e-7;
  const long long N = 5e5 + 7, mod = 1e9 + 7;
  long long POW(long long a, long long b, long long mod) {
    a %= mod;
    long long r = 1;
    for (long long i = b; i > 0; i >>= 1) {
      if (i & 1) r = (r * a) % mod;
      a = (a * a) % mod;
    }
    return r;
  }
  long long f[N];
  void init() {
    f[0] = 1;
    for (long long i = 1; i < N; i++) {
      f[i] = (f[i - 1] * i) % mod;
    }
  }
  long long nCr(long long n, long long r) {
    if (n < r) return 0;
    return f[n] * POW(f[n - r] * f[r], mod - 2, mod) % mod;
  }
  long long nPr(long long n, long long r) { return nCr(n, r) * f[r] % mod; }
  long long p = 1e9 + 7;
  long long modadd(long long a, long long b) {
    a = a % p;
    b = b % p;
    return (((a + b) % p) + p) % p;
  }
  long long modmul(long long a, long long b) {
    a = a % p;
    b = b % p;
    return (((a * b) % p) + p) % p;
  }
  long long modsub(long long a, long long b) {
    a = a % p;
    b = b % p;
    return (((a - b) % p) + p) % p;
  }
  void solve() {
    long long n;
    cin >> n;
    long long arr[n];
    for (int i = 0; i < n; i += 1) cin >> arr[i];
    map<long long, long long> m;
    long long f = 0;
    for (int i = 0; i < n; i += 1) {
      m[arr[i]]++;
      if (m[arr[i]] > 1) {
        f = 1;
        break;
      }
    }
    if (f == 1) {
      cout << "YES" << endl;
      return;
    }
    bool vis[n + 1];
    fill(vis, vis + n, 0);
    long long c = 0;
    for (int i = 0; i < n; i += 1) {
      if (!vis[i]) {
        long long s = 1;
        vis[i] = 1;
        long long t = arr[i];
        while (1) {
          if (!vis[t - 1]) {
            s++;
            vis[t - 1] = 1;
            t = arr[t - 1];
          } else
            break;
        }
        if (s % 2 == 0) c++;
      }
    }
    if (c % 2 == 0)
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
    return;
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  