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 sz = 5e5 + 10;
  long long y[sz];
  inline long long read() {
    long long f = 1, x = 0;
    char ch = getchar();
    for (; !isdigit(ch); ch = getchar())
      if (ch == '-') f = -1;
    for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
    return f * x;
  }
  long long ans, n;
  long long c[sz], a[sz];
  void msort(long long b, long long e) {
    if (b == e) return;
    long long mid = (b + e) / 2, i = b, j = mid + 1, k = b;
    msort(b, mid), msort(mid + 1, e);
    while (i <= mid && j <= e)
      if (a[i] <= a[j])
        c[k++] = a[i++];
      else
        c[k++] = a[j++], ans += mid - i + 1;
    while (i <= mid) c[k++] = a[i++];
    while (j <= e) c[k++] = a[j++];
    for (long long l = b; l <= e; l++) a[l] = c[l];
  }
  bool z[sz];
  void solve() {
    long long n = read();
    memset(z, 0, sizeof(bool) * (n + 2));
    bool f = 0;
    for (long long i = 1; i <= n; i++) {
      a[i] = read();
      if (!z[a[i]])
        z[a[i]] = 1;
      else
        f = 1;
    }
    ans = 0;
    msort(1, n);
    if (ans & 1) {
      if (f)
        puts("YES");
      else
        puts("NO");
    } else
      puts("YES");
  }
  signed main() {
    long long t;
    cin >> t;
    while (t--) {
      solve();
    }
  }
  