#include <bits/stdc++.h>
using namespace std;
template <typename T>
struct binary_indexed_tree {
  int N;
  vector<T> BIT;
  binary_indexed_tree(int N) : N(N), BIT(N + 1, 0) {}
  void add(int i, T x) {
    i++;
    while (i <= N) {
      BIT[i] += x;
      i += i & -i;
    }
  }
  T sum(int i) {
    T ans = 0;
    while (i > 0) {
      ans += BIT[i];
      i -= i & -i;
    }
    return ans;
  }
  T sum(int L, int R) { return sum(R) - sum(L); }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int j = 0; j < n; j++) {
      cin >> a[j];
      a[j]--;
    }
    binary_indexed_tree<int> BIT(n);
    long long S = 0;
    for (int j = 0; j < n; j++) {
      S += BIT.sum(a[j] + 1, n);
      BIT.add(a[j], 1);
    }
    if (S % 2 == 0) {
      cout << "YES"
           << "\n";
    } else {
      sort(a.begin(), a.end());
      bool ok = false;
      for (int j = 0; j < n - 1; j++) {
        if (a[j] == a[j + 1]) {
          ok = true;
        }
      }
      if (ok) {
        cout << "YES"
             << "\n";
      } else {
        cout << "NO"
             << "\n";
      }
    }
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
template <typename T>
struct binary_indexed_tree {
  int N;
  vector<T> BIT;
  binary_indexed_tree(int N) : N(N), BIT(N + 1, 0) {}
  void add(int i, T x) {
    i++;
    while (i <= N) {
      BIT[i] += x;
      i += i & -i;
    }
  }
  T sum(int i) {
    T ans = 0;
    while (i > 0) {
      ans += BIT[i];
      i -= i & -i;
    }
    return ans;
  }
  T sum(int L, int R) { return sum(R) - sum(L); }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int j = 0; j < n; j++) {
      cin >> a[j];
      a[j]--;
    }
    binary_indexed_tree<int> BIT(n);
    long long S = 0;
    for (int j = 0; j < n; j++) {
      S += BIT.sum(a[j] + 1, n);
      BIT.add(a[j], 1);
    }
    if (S % 2 == 0) {
      cout << "YES"
           << "\n";
    } else {
      sort(a.begin(), a.end());
      bool ok = false;
      for (int j = 0; j < n - 1; j++) {
        if (a[j] == a[j + 1]) {
          ok = true;
        }
      }
      if (ok) {
        cout << "YES"
             << "\n";
      } else {
        cout << "NO"
             << "\n";
      }
    }
  }
}

```