#include <bits/stdc++.h>
using namespace std;
const long long NMX = 432000, oo = 0x3C3C3C3C3C3C3C3C, Bs = 1000000007;
long long q, n, a[NMX], A[NMX], kq;
bool dd[NMX], YES;
void update(long long x) {
  while (x > 0) {
    A[x]++;
    x -= ((x) & (-(x)));
  }
}
long long get(long long x) {
  long long res = 0;
  while (x <= n) {
    res += A[x];
    x += ((x) & (-(x)));
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> q;
  while (q--) {
    cin >> n;
    for (int i = 1; i <= (int)(n); i++) {
      cin >> a[i];
      A[i] = 0;
      dd[i] = false;
    }
    YES = false;
    for (int i = 1; i <= (int)(n); i++) {
      if (dd[a[i]]) {
        cout << "YES\n";
        YES = true;
        break;
      }
      dd[a[i]] = true;
    }
    if (YES) continue;
    kq = 0;
    for (int i = 1; i <= (int)(n); i++) {
      kq += get(a[i] + 1);
      update(a[i]);
    }
    cout << (kq % 2 == 0 ? "YES\n" : "NO\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 long long NMX = 432000, oo = 0x3C3C3C3C3C3C3C3C, Bs = 1000000007;
long long q, n, a[NMX], A[NMX], kq;
bool dd[NMX], YES;
void update(long long x) {
  while (x > 0) {
    A[x]++;
    x -= ((x) & (-(x)));
  }
}
long long get(long long x) {
  long long res = 0;
  while (x <= n) {
    res += A[x];
    x += ((x) & (-(x)));
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> q;
  while (q--) {
    cin >> n;
    for (int i = 1; i <= (int)(n); i++) {
      cin >> a[i];
      A[i] = 0;
      dd[i] = false;
    }
    YES = false;
    for (int i = 1; i <= (int)(n); i++) {
      if (dd[a[i]]) {
        cout << "YES\n";
        YES = true;
        break;
      }
      dd[a[i]] = true;
    }
    if (YES) continue;
    kq = 0;
    for (int i = 1; i <= (int)(n); i++) {
      kq += get(a[i] + 1);
      update(a[i]);
    }
    cout << (kq % 2 == 0 ? "YES\n" : "NO\n");
  }
  return 0;
}

```