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 MAXN = 5e5 + 7;
  const int maxn = 5e4 + 7;
  const int inf = 1e6 + 10;
  const double pp = 1e-6;
  template <class type>
  inline void read(type &k) {
    k = 0;
    type t = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') t = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      k = k * 10 + ch - '0';
      ch = getchar();
    }
    k *= t;
  }
  inline long long gcd(long long x, long long y) {
    return y == 0 ? x : gcd(y, x % y);
  }
  long long qpow(long long a, long long b) {
    long long aa = 1;
    while (b) {
      if (b % 2 == 1) aa = (aa * a);
      a = (a * a);
      b /= 2;
    }
    return aa;
  }
  int a[MAXN], b[MAXN], tmp[MAXN];
  long long merge_sort(int q[], int l, int r) {
    if (l >= r) return 0;
    int mid = l + r >> 1;
    long long res = merge_sort(q, l, mid) + merge_sort(q, mid + 1, r);
    int k = 0, i = l, j = mid + 1;
    while (i <= mid && j <= r)
      if (q[i] <= q[j])
        tmp[k++] = q[i++];
      else {
        res += mid - i + 1;
        tmp[k++] = q[j++];
      }
    while (i <= mid) tmp[k++] = q[i++];
    while (j <= r) tmp[k++] = q[j++];
    for (i = l, j = 0; i <= r; i++, j++) q[i] = tmp[j];
    return res;
  }
  int main() {
    int t;
    read(t);
    while (t--) {
      int n;
      bool flag = false;
      read(n);
      for (int i = 1; i <= n; i++) {
        read(a[i]);
        b[i] = a[i];
      }
      sort(b + 1, b + 1 + n);
      int now = 1;
      for (int i = 1; i <= n; i++) {
        if (i != 1 && b[i] == b[i - 1]) flag = true;
      }
      int ans = merge_sort(a, 1, n);
      printf((ans % 2) && !flag ? "NO\n" : "YES\n");
    }
  }
  