#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const long long INF = 1e9;
const long double EPS = 1e-9;
const long long N = 1e6 + 2;
long long mod(long long x, long long y = MOD) { return ((x % y) + y) % y; }
long long addm(long long x) { return x; }
template <typename... Ts>
long long addm(long long x, Ts... ys) {
  return mod(x + addm(ys...));
}
long long mulm(long long x) { return x; }
template <typename... Ts>
long long mulm(long long x, Ts... ys) {
  return mod(x * mulm(ys...));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    pair<long long, long long> arr[n];
    vector<long long> loop;
    int visited[n];
    memset(visited, 0, sizeof visited);
    ;
    for (long long i = (0); i < (n); i++) {
      int a;
      cin >> a;
      arr[i] = pair<long long, long long>(a, i);
    }
    sort(arr, arr + n);
    bool flag = false;
    for (long long i = (1); i < (n); i++) {
      if (arr[i].first == arr[i - 1].first) flag = true;
    }
    if (flag) {
      cout << "YES\n";
      continue;
    }
    int count = 0;
    int oddcount = 0;
    for (long long i = (0); i < (n); i++) {
      if (visited[i]) continue;
      int u = i;
      int cnt = 0;
      while (!visited[u]) {
        cnt++;
        visited[u] = 1;
        u = arr[u].second;
      }
      if (cnt % 2 == 0)
        count++;
      else if (cnt != 1)
        oddcount++;
    }
    if (count % 2 == 1)
      cout << "NO\n";
    else
      cout << "YES\n";
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 long long MOD = 1e9 + 7;
const long long INF = 1e9;
const long double EPS = 1e-9;
const long long N = 1e6 + 2;
long long mod(long long x, long long y = MOD) { return ((x % y) + y) % y; }
long long addm(long long x) { return x; }
template <typename... Ts>
long long addm(long long x, Ts... ys) {
  return mod(x + addm(ys...));
}
long long mulm(long long x) { return x; }
template <typename... Ts>
long long mulm(long long x, Ts... ys) {
  return mod(x * mulm(ys...));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    pair<long long, long long> arr[n];
    vector<long long> loop;
    int visited[n];
    memset(visited, 0, sizeof visited);
    ;
    for (long long i = (0); i < (n); i++) {
      int a;
      cin >> a;
      arr[i] = pair<long long, long long>(a, i);
    }
    sort(arr, arr + n);
    bool flag = false;
    for (long long i = (1); i < (n); i++) {
      if (arr[i].first == arr[i - 1].first) flag = true;
    }
    if (flag) {
      cout << "YES\n";
      continue;
    }
    int count = 0;
    int oddcount = 0;
    for (long long i = (0); i < (n); i++) {
      if (visited[i]) continue;
      int u = i;
      int cnt = 0;
      while (!visited[u]) {
        cnt++;
        visited[u] = 1;
        u = arr[u].second;
      }
      if (cnt % 2 == 0)
        count++;
      else if (cnt != 1)
        oddcount++;
    }
    if (count % 2 == 1)
      cout << "NO\n";
    else
      cout << "YES\n";
  }
  return 0;
}

```