#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    vector<int> a_ = a;
    sort(a.begin(), a.end());
    bool f = false;
    for (int i = 0; i < n - 1; i++) {
      if (a[i] == a[i + 1]) {
        f = true;
        break;
      }
    }
    if (!f) {
      vector<pair<int, int>> pos(n);
      for (int i = 0; i < n; i++) {
        pos[i] = {a_[i], i};
      }
      sort(pos.begin(), pos.end());
      vector<bool> used(n, false);
      int ans = 0;
      for (int i = 0; i < n; i++) {
        if (!used[i]) {
          int cnt = 0;
          int start = i;
          int curr = i;
          while (true) {
            curr = pos[curr].second;
            cnt++;
            used[curr] = true;
            if (curr == start) {
              break;
            }
          }
          ans += cnt - 1;
        }
      }
      if (ans % 2 == 0) {
        cout << "YES\n";
      } else {
        cout << "NO\n";
      }
    } else {
      cout << "YES\n";
    }
  }
  return 0;
}

### Prompt

Please formulate 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;
using ll = long long;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    vector<int> a_ = a;
    sort(a.begin(), a.end());
    bool f = false;
    for (int i = 0; i < n - 1; i++) {
      if (a[i] == a[i + 1]) {
        f = true;
        break;
      }
    }
    if (!f) {
      vector<pair<int, int>> pos(n);
      for (int i = 0; i < n; i++) {
        pos[i] = {a_[i], i};
      }
      sort(pos.begin(), pos.end());
      vector<bool> used(n, false);
      int ans = 0;
      for (int i = 0; i < n; i++) {
        if (!used[i]) {
          int cnt = 0;
          int start = i;
          int curr = i;
          while (true) {
            curr = pos[curr].second;
            cnt++;
            used[curr] = true;
            if (curr == start) {
              break;
            }
          }
          ans += cnt - 1;
        }
      }
      if (ans % 2 == 0) {
        cout << "YES\n";
      } else {
        cout << "NO\n";
      }
    } else {
      cout << "YES\n";
    }
  }
  return 0;
}

```