#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
long long merge(long long arr[], long long l, long long mid, long long r) {
  long long ans = 0;
  long long n1 = mid - l + 1;
  long long n2 = r - mid;
  long long a[n1];
  long long b[n2];
  for (long long i = 0; i < n1; i++) {
    a[i] = arr[l + i];
  }
  for (long long i = 0; i < n2; i++) {
    b[i] = arr[mid + 1 + i];
  }
  long long i = 0;
  long long j = 0;
  long long k = l;
  while (i < n1 && j < n2) {
    if (a[i] <= b[j]) {
      arr[k] = a[i];
      k++;
      i++;
    } else {
      ans += n1 - i;
      arr[k] = b[j];
      k++;
      j++;
    }
  }
  while (i < n1) {
    arr[k] = a[i];
    k++;
    i++;
  }
  while (j < n2) {
    arr[k] = b[j];
    k++;
    j++;
  }
  return ans;
}
long long inversion_count(long long arr[], long long l, long long r) {
  long long ans = 0;
  if (l < r) {
    long long mid = (l + r) / 2;
    ans += inversion_count(arr, l, mid);
    ans += inversion_count(arr, mid + 1, r);
    ans += merge(arr, l, mid, r);
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long v[n];
    for (long long i = 0; i < n; i++) cin >> v[i];
    bool ans = false;
    long long inv = inversion_count(v, 0, n - 1);
    if (inv % 2 == 0)
      ans = true;
    else {
      sort(v, v + n);
      for (long long i = 0; i < n - 1; i++)
        if (v[i] == v[i + 1]) ans = true;
    }
    cout << (ans ? "YES" : "NO") << '\n';
  }
  return 0;
}

### Prompt

Please create a solution in Cpp to 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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
long long merge(long long arr[], long long l, long long mid, long long r) {
  long long ans = 0;
  long long n1 = mid - l + 1;
  long long n2 = r - mid;
  long long a[n1];
  long long b[n2];
  for (long long i = 0; i < n1; i++) {
    a[i] = arr[l + i];
  }
  for (long long i = 0; i < n2; i++) {
    b[i] = arr[mid + 1 + i];
  }
  long long i = 0;
  long long j = 0;
  long long k = l;
  while (i < n1 && j < n2) {
    if (a[i] <= b[j]) {
      arr[k] = a[i];
      k++;
      i++;
    } else {
      ans += n1 - i;
      arr[k] = b[j];
      k++;
      j++;
    }
  }
  while (i < n1) {
    arr[k] = a[i];
    k++;
    i++;
  }
  while (j < n2) {
    arr[k] = b[j];
    k++;
    j++;
  }
  return ans;
}
long long inversion_count(long long arr[], long long l, long long r) {
  long long ans = 0;
  if (l < r) {
    long long mid = (l + r) / 2;
    ans += inversion_count(arr, l, mid);
    ans += inversion_count(arr, mid + 1, r);
    ans += merge(arr, l, mid, r);
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long v[n];
    for (long long i = 0; i < n; i++) cin >> v[i];
    bool ans = false;
    long long inv = inversion_count(v, 0, n - 1);
    if (inv % 2 == 0)
      ans = true;
    else {
      sort(v, v + n);
      for (long long i = 0; i < n - 1; i++)
        if (v[i] == v[i + 1]) ans = true;
    }
    cout << (ans ? "YES" : "NO") << '\n';
  }
  return 0;
}

```