#include <bits/stdc++.h>
using namespace std;
const int maxn = 5e5 + 5;
int d[maxn];
unordered_map<int, int> mp;
struct node {
  int id;
  int va;
};
vector<node> vec;
bool cmp(node a, node b) { return a.va < b.va; }
bool vis[maxn];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    vec.clear();
    mp.clear();
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) {
      int x;
      cin >> x;
      mp[x]++;
      vis[i] = 0;
      vec.push_back({i, x});
    }
    bool flag = false;
    for (auto i : mp) {
      if (i.second >= 2) {
        cout << "YES" << '\n';
        ;
        flag = true;
        break;
      }
    }
    if (flag) continue;
    sort(vec.begin(), vec.end(), cmp);
    for (int i = 0; i < n; i++) {
      d[vec[i].id] = i + 1;
    }
    int cnt = 0;
    for (int i = 1; i <= n; i++) {
      if (vis[i]) continue;
      int to = i;
      vis[to] = true;
      to = d[to];
      int ta = 1;
      while (to != i) {
        vis[to] = true;
        ta++;
        to = d[to];
      }
      if (ta & 1)
        ;
      else
        cnt++;
    }
    cout << ((cnt & 1) ? "NO" : "YES") << '\n';
    ;
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
const int maxn = 5e5 + 5;
int d[maxn];
unordered_map<int, int> mp;
struct node {
  int id;
  int va;
};
vector<node> vec;
bool cmp(node a, node b) { return a.va < b.va; }
bool vis[maxn];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    vec.clear();
    mp.clear();
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) {
      int x;
      cin >> x;
      mp[x]++;
      vis[i] = 0;
      vec.push_back({i, x});
    }
    bool flag = false;
    for (auto i : mp) {
      if (i.second >= 2) {
        cout << "YES" << '\n';
        ;
        flag = true;
        break;
      }
    }
    if (flag) continue;
    sort(vec.begin(), vec.end(), cmp);
    for (int i = 0; i < n; i++) {
      d[vec[i].id] = i + 1;
    }
    int cnt = 0;
    for (int i = 1; i <= n; i++) {
      if (vis[i]) continue;
      int to = i;
      vis[to] = true;
      to = d[to];
      int ta = 1;
      while (to != i) {
        vis[to] = true;
        ta++;
        to = d[to];
      }
      if (ta & 1)
        ;
      else
        cnt++;
    }
    cout << ((cnt & 1) ? "NO" : "YES") << '\n';
    ;
  }
  return 0;
}

```