#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const long long N = 2e5 + 10;
long long ans = 0;
void mergeSort(vector<long long> &v, int st, int end) {
  if (st == end) {
    return;
  }
  int mid = (st + end) / 2;
  mergeSort(v, st, mid);
  mergeSort(v, mid + 1, end);
  vector<int> a, b;
  for (int i = st; i <= mid; i++) {
    a.push_back(v[i]);
  }
  for (int i = mid + 1; i <= end; i++) {
    b.push_back(v[i]);
  }
  int i = 0, j = 0, curr = st;
  int aa = a.size(), bb = b.size();
  while (i < aa or j < bb) {
    if (i < aa and j < bb and a[i] <= b[j]) {
      v[curr++] = a[i++];
    } else if (i < aa and j < bb and a[i] > b[j]) {
      v[curr++] = b[j++];
      ans += aa - i;
    } else if (i < aa) {
      v[curr++] = a[i++];
    } else {
      v[curr++] = b[j++];
    }
  }
}
long long countInversions(vector<long long> &v, int n) {
  mergeSort(v, 0, n - 1);
  return ans;
}
void test_case() {
  long long n;
  cin >> n;
  vector<long long> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }
  ans = 0;
  countInversions(v, n);
  set<long long> st(v.begin(), v.end());
  if (st.size() < n or ans % 2 == 0) {
    cout << "YES\n";
  } else {
    cout << "NO\n";
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int tc = 1;
  cin >> tc;
  while (tc--) {
    test_case();
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
const long long MOD = 1e9 + 7;
const long long N = 2e5 + 10;
long long ans = 0;
void mergeSort(vector<long long> &v, int st, int end) {
  if (st == end) {
    return;
  }
  int mid = (st + end) / 2;
  mergeSort(v, st, mid);
  mergeSort(v, mid + 1, end);
  vector<int> a, b;
  for (int i = st; i <= mid; i++) {
    a.push_back(v[i]);
  }
  for (int i = mid + 1; i <= end; i++) {
    b.push_back(v[i]);
  }
  int i = 0, j = 0, curr = st;
  int aa = a.size(), bb = b.size();
  while (i < aa or j < bb) {
    if (i < aa and j < bb and a[i] <= b[j]) {
      v[curr++] = a[i++];
    } else if (i < aa and j < bb and a[i] > b[j]) {
      v[curr++] = b[j++];
      ans += aa - i;
    } else if (i < aa) {
      v[curr++] = a[i++];
    } else {
      v[curr++] = b[j++];
    }
  }
}
long long countInversions(vector<long long> &v, int n) {
  mergeSort(v, 0, n - 1);
  return ans;
}
void test_case() {
  long long n;
  cin >> n;
  vector<long long> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }
  ans = 0;
  countInversions(v, n);
  set<long long> st(v.begin(), v.end());
  if (st.size() < n or ans % 2 == 0) {
    cout << "YES\n";
  } else {
    cout << "NO\n";
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int tc = 1;
  cin >> tc;
  while (tc--) {
    test_case();
  }
  return 0;
}

```