#include <bits/stdc++.h>
using namespace std;
int arr[500005], brr[500005];
pair<int, int> parr[500005];
int ord[500005];
int dfs(int idx) {
  if (ord[idx] != 0) return 0;
  int ret = 1;
  vector<int> same_ord;
  int nex = idx;
  while (brr[nex] != idx) {
    nex = brr[nex];
    same_ord.push_back(nex);
    ret += 1;
  }
  for (int i = 0; i < same_ord.size(); i++) {
    ord[same_ord[i]] = ret;
  }
  ord[idx] = ret;
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
      cin >> parr[i].first;
      parr[i].second = i;
      ord[i + 1] = 0;
    }
    sort(parr, parr + n);
    int ev = 0;
    for (int i = 0; i < n - 1; i++) {
      if (parr[i].first == parr[i + 1].first) {
        goto YES;
      }
    }
    for (int i = 0; i < n; i++) {
      arr[i + 1] = parr[i].second + 1;
    }
    for (int i = 1; i <= n; i++) {
      brr[arr[i]] = i;
    }
    for (int i = 1; i < n + 1; i++) {
      int x = dfs(i);
      if (x == 0)
        continue;
      else {
        if (x % 2 == 0) ev += 1;
      }
    }
  YES:
    if (ev % 2 == 0)
      cout << "YES" << '\n';
    else
      cout << "NO" << '\n';
  }
}

### 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;
int arr[500005], brr[500005];
pair<int, int> parr[500005];
int ord[500005];
int dfs(int idx) {
  if (ord[idx] != 0) return 0;
  int ret = 1;
  vector<int> same_ord;
  int nex = idx;
  while (brr[nex] != idx) {
    nex = brr[nex];
    same_ord.push_back(nex);
    ret += 1;
  }
  for (int i = 0; i < same_ord.size(); i++) {
    ord[same_ord[i]] = ret;
  }
  ord[idx] = ret;
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
      cin >> parr[i].first;
      parr[i].second = i;
      ord[i + 1] = 0;
    }
    sort(parr, parr + n);
    int ev = 0;
    for (int i = 0; i < n - 1; i++) {
      if (parr[i].first == parr[i + 1].first) {
        goto YES;
      }
    }
    for (int i = 0; i < n; i++) {
      arr[i + 1] = parr[i].second + 1;
    }
    for (int i = 1; i <= n; i++) {
      brr[arr[i]] = i;
    }
    for (int i = 1; i < n + 1; i++) {
      int x = dfs(i);
      if (x == 0)
        continue;
      else {
        if (x % 2 == 0) ev += 1;
      }
    }
  YES:
    if (ev % 2 == 0)
      cout << "YES" << '\n';
    else
      cout << "NO" << '\n';
  }
}

```