#include <bits/stdc++.h>
const long long N = 2e5 + 1;
using namespace std;
clock_t startTime;
double getCurrentTime() {
  return (double)(clock() - startTime) / CLOCKS_PER_SEC;
}
void init_code() {}
signed main() {
  startTime = clock();
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long T = 1;
  cin >> T;
  while (T--) {
    long long n;
    cin >> n;
    vector<long long> v(n);
    map<long long, long long> mp;
    for (long long i = 0; i < n; i++) {
      cin >> v[i];
      mp[v[i]] = i;
    }
    if (mp.size() < n) {
      cout << "YES"
           << "\n";
      continue;
    }
    vector<long long> temp = v;
    sort(temp.begin(), temp.end());
    long long swaps = 0;
    for (long long i = 0; i < n; i++) {
      if (temp[i] != v[i]) {
        swaps++;
        long long pos = mp[temp[i]];
        mp[temp[i]] = i;
        mp[v[i]] = pos;
        swap(v[i], v[pos]);
      }
    }
    if (swaps % 2)
      cout << "NO";
    else
      cout << "YES";
    cout << "\n";
  }
}

### Prompt

Generate 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>
const long long N = 2e5 + 1;
using namespace std;
clock_t startTime;
double getCurrentTime() {
  return (double)(clock() - startTime) / CLOCKS_PER_SEC;
}
void init_code() {}
signed main() {
  startTime = clock();
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long T = 1;
  cin >> T;
  while (T--) {
    long long n;
    cin >> n;
    vector<long long> v(n);
    map<long long, long long> mp;
    for (long long i = 0; i < n; i++) {
      cin >> v[i];
      mp[v[i]] = i;
    }
    if (mp.size() < n) {
      cout << "YES"
           << "\n";
      continue;
    }
    vector<long long> temp = v;
    sort(temp.begin(), temp.end());
    long long swaps = 0;
    for (long long i = 0; i < n; i++) {
      if (temp[i] != v[i]) {
        swaps++;
        long long pos = mp[temp[i]];
        mp[temp[i]] = i;
        mp[v[i]] = pos;
        swap(v[i], v[pos]);
      }
    }
    if (swaps % 2)
      cout << "NO";
    else
      cout << "YES";
    cout << "\n";
  }
}

```