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 NR = 5e5 + 10;
  int T, n;
  int a[NR], num[NR];
  struct arr {
    int mx, pos;
  } t[NR * 4];
  void build(int x, int lx, int rx) {
    if (lx == rx) {
      t[x].mx = a[lx];
      t[x].pos = lx;
      return;
    }
    int mid = (lx + rx) >> 1;
    build(x * 2, lx, mid);
    build(x * 2 + 1, mid + 1, rx);
    if (t[x * 2].mx > t[x * 2 + 1].mx)
      t[x] = t[x * 2];
    else
      t[x] = t[x * 2 + 1];
  }
  void add(int x, int lx, int rx, int p, int k) {
    if (lx == rx) {
      t[x].mx = k;
      t[x].pos = lx;
      return;
    }
    int mid = (lx + rx) >> 1;
    if (p <= mid)
      add(x * 2, lx, mid, p, k);
    else
      add(x * 2 + 1, mid + 1, rx, p, k);
    if (t[x * 2].mx > t[x * 2 + 1].mx)
      t[x] = t[x * 2];
    else
      t[x] = t[x * 2 + 1];
  }
  int main() {
    scanf("%d", &T);
    while (T--) {
      int flag = 0;
      scanf("%d", &n);
      for (int i = 1; i <= n; i++) num[i] = 0;
      for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        num[a[i]]++;
        if (num[a[i]] >= 2) flag = 1;
      }
      if (flag) {
        printf("YES\n");
        continue;
      }
      build(1, 1, n);
      flag = 1;
      for (int i = n; i >= 1; i--) {
        int x = t[1].pos, y = 1;
        if (x == i) {
          add(1, 1, n, i, 0);
          continue;
        }
        if (x == 1 && i == 1) break;
        if (x == 1 && i == 2) {
          flag = 0;
          break;
        }
        if (x == 1) y = 2;
        int tmp = a[x];
        add(1, 1, n, x, a[y]);
        add(1, 1, n, y, a[i]);
        add(1, 1, n, i, 0);
        a[x] = a[y];
        a[y] = a[i];
        a[i] = tmp;
      }
      if (flag)
        printf("YES\n");
      else
        printf("NO\n");
    }
    return 0;
  }
  