#include <bits/stdc++.h>
using namespace std;
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  vector<pair<int, int>> b(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    b[i].first = a[i];
    b[i].second = i;
  }
  sort(b.begin(), b.end());
  for (int i = 1; i < n; i++) {
    if (b[i].first == b[i - 1].first) {
      cout << "YES\n";
      return;
    }
  }
  vector<int> pos(n);
  for (int i = 0; i < n; i++) {
    a[b[i].second] = i;
  }
  for (int i = 0; i < n; i++) {
    pos[a[i]] = i;
  }
  for (int i = 0; i < n - 2; i++) {
    if (a[i] == i)
      continue;
    else if (a[i] != n - 1 && a[i] != n - 2) {
      pos[a[i]] = pos[n - 1];
      pos[n - 1] = pos[i];
      a[pos[n - 1]] = n - 1;
      a[pos[a[i]]] = a[i];
    } else if (a[i] == n - 1) {
      pos[n - 1] = pos[n - 2];
      pos[n - 2] = pos[i];
      a[pos[n - 1]] = n - 1;
      a[pos[n - 2]] = n - 2;
    } else if (a[i] == n - 2) {
      pos[n - 2] = pos[n - 1];
      pos[n - 1] = pos[i];
      a[pos[n - 1]] = n - 1;
      a[pos[n - 2]] = n - 2;
    }
  }
  if (pos[n - 1] == n - 1) {
    cout << "YES\n";
  } else
    cout << "NO\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) solve();
}

### Prompt

Your task is to create a Cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  vector<pair<int, int>> b(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    b[i].first = a[i];
    b[i].second = i;
  }
  sort(b.begin(), b.end());
  for (int i = 1; i < n; i++) {
    if (b[i].first == b[i - 1].first) {
      cout << "YES\n";
      return;
    }
  }
  vector<int> pos(n);
  for (int i = 0; i < n; i++) {
    a[b[i].second] = i;
  }
  for (int i = 0; i < n; i++) {
    pos[a[i]] = i;
  }
  for (int i = 0; i < n - 2; i++) {
    if (a[i] == i)
      continue;
    else if (a[i] != n - 1 && a[i] != n - 2) {
      pos[a[i]] = pos[n - 1];
      pos[n - 1] = pos[i];
      a[pos[n - 1]] = n - 1;
      a[pos[a[i]]] = a[i];
    } else if (a[i] == n - 1) {
      pos[n - 1] = pos[n - 2];
      pos[n - 2] = pos[i];
      a[pos[n - 1]] = n - 1;
      a[pos[n - 2]] = n - 2;
    } else if (a[i] == n - 2) {
      pos[n - 2] = pos[n - 1];
      pos[n - 1] = pos[i];
      a[pos[n - 1]] = n - 1;
      a[pos[n - 2]] = n - 2;
    }
  }
  if (pos[n - 1] == n - 1) {
    cout << "YES\n";
  } else
    cout << "NO\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) solve();
}

```