#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> v(n);
    vector<int> nxt(n);
    for (int &i : v) {
      cin >> i;
    }
    vector<pair<int, int>> v1(n);
    for (int i = 0; i < n; i++) {
      v1[i] = make_pair(v[i], i);
    }
    sort(v1.begin(), v1.end());
    for (int i = 0; i < n; i++) {
      nxt[v1[i].second] = i;
    }
    bool same = false;
    for (int i = 0; i < n - 1; i++) {
      if (v1[i].first == v1[i + 1].first) same = true;
    }
    vector<int> vis(n);
    int countcycle = 0;
    for (int i = 0; i < n; i++) {
      if (vis[i]) continue;
      if (nxt[i] != i) {
        int countlen = 0;
        while (nxt[i] != i && !vis[i]) {
          vis[i] = 1;
          countlen++;
          i = nxt[i];
        }
        if (countlen % 2 == 0) countcycle++;
      }
    }
    cout << (((countcycle % 2 == 0) || same) ? "YES" : "NO") << endl;
  }
  return 0;
}

### 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;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> v(n);
    vector<int> nxt(n);
    for (int &i : v) {
      cin >> i;
    }
    vector<pair<int, int>> v1(n);
    for (int i = 0; i < n; i++) {
      v1[i] = make_pair(v[i], i);
    }
    sort(v1.begin(), v1.end());
    for (int i = 0; i < n; i++) {
      nxt[v1[i].second] = i;
    }
    bool same = false;
    for (int i = 0; i < n - 1; i++) {
      if (v1[i].first == v1[i + 1].first) same = true;
    }
    vector<int> vis(n);
    int countcycle = 0;
    for (int i = 0; i < n; i++) {
      if (vis[i]) continue;
      if (nxt[i] != i) {
        int countlen = 0;
        while (nxt[i] != i && !vis[i]) {
          vis[i] = 1;
          countlen++;
          i = nxt[i];
        }
        if (countlen % 2 == 0) countcycle++;
      }
    }
    cout << (((countcycle % 2 == 0) || same) ? "YES" : "NO") << endl;
  }
  return 0;
}

```