#include <bits/stdc++.h>
using namespace std;
vector<long long> *adj;
bool *visited;
void solve() {
  long long n;
  cin >> n;
  vector<long long> v(n);
  unordered_set<long long> s;
  bool as = 1;
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
    if (s.find(v[i]) != s.end()) {
      as = 0;
    } else {
      s.insert(v[i]);
    }
  }
  if (as == 0) {
    cout << "YES" << endl;
  } else {
    vector<long long> u = v;
    sort(u.begin(), u.end());
    unordered_map<long long, long long> m;
    vector<bool> b(n, 0);
    for (long long i = 0; i < n; i++) {
      m[u[i]] = i;
    }
    vector<long long> ans;
    for (long long i = 0; i < n; i++) {
      if (b[i] == 0) {
        long long j = i;
        long long c = 0;
        b[j] = 1;
        while (1) {
          j = m[v[j]];
          c++;
          b[j] = 1;
          if (j == i) {
            break;
          }
        }
        ans.push_back(c);
      }
    }
    long long tot = 0;
    for (long long i = 0; i < ans.size(); i++) {
      if (ans[i] % 2 == 0) {
        tot++;
      }
    }
    if (tot % 2 == 0) {
      cout << "YES" << endl;
    } else {
      cout << "NO" << endl;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
vector<long long> *adj;
bool *visited;
void solve() {
  long long n;
  cin >> n;
  vector<long long> v(n);
  unordered_set<long long> s;
  bool as = 1;
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
    if (s.find(v[i]) != s.end()) {
      as = 0;
    } else {
      s.insert(v[i]);
    }
  }
  if (as == 0) {
    cout << "YES" << endl;
  } else {
    vector<long long> u = v;
    sort(u.begin(), u.end());
    unordered_map<long long, long long> m;
    vector<bool> b(n, 0);
    for (long long i = 0; i < n; i++) {
      m[u[i]] = i;
    }
    vector<long long> ans;
    for (long long i = 0; i < n; i++) {
      if (b[i] == 0) {
        long long j = i;
        long long c = 0;
        b[j] = 1;
        while (1) {
          j = m[v[j]];
          c++;
          b[j] = 1;
          if (j == i) {
            break;
          }
        }
        ans.push_back(c);
      }
    }
    long long tot = 0;
    for (long long i = 0; i < ans.size(); i++) {
      if (ans[i] % 2 == 0) {
        tot++;
      }
    }
    if (tot % 2 == 0) {
      cout << "YES" << endl;
    } else {
      cout << "NO" << endl;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```