#include <bits/stdc++.h>
using namespace std;
const int N = 5e5 + 10;
int a[N];
int c[N];
int tree[N];
int lowbit(int x) { return x & -x; }
int sum(int x) {
  int res = 0;
  for (int i = x; i; i -= lowbit(i)) res += tree[i];
  return res;
}
void add(int x, int n) {
  for (int i = x; i <= n; i += lowbit(i)) tree[i]++;
}
void solve() {
  int n;
  cin >> n;
  memset(c, 0, (n + 2) * sizeof(int));
  memset(tree, 0, (n + 2) * sizeof(int));
  int maxx = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    c[a[i]]++;
    maxx = max(maxx, c[a[i]]);
  }
  if (maxx > 1) {
    cout << "YES" << endl;
    return;
  }
  int res = 0;
  for (int i = 0; i < n; i++) {
    res += sum(n) - sum(a[i]);
    ;
    add(a[i], n);
  }
  if (res & 1)
    cout << "NO" << endl;
  else
    cout << "YES" << endl;
}
int main() {
  int t;
  cin >> t;
  while (t--) solve();
  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 N = 5e5 + 10;
int a[N];
int c[N];
int tree[N];
int lowbit(int x) { return x & -x; }
int sum(int x) {
  int res = 0;
  for (int i = x; i; i -= lowbit(i)) res += tree[i];
  return res;
}
void add(int x, int n) {
  for (int i = x; i <= n; i += lowbit(i)) tree[i]++;
}
void solve() {
  int n;
  cin >> n;
  memset(c, 0, (n + 2) * sizeof(int));
  memset(tree, 0, (n + 2) * sizeof(int));
  int maxx = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    c[a[i]]++;
    maxx = max(maxx, c[a[i]]);
  }
  if (maxx > 1) {
    cout << "YES" << endl;
    return;
  }
  int res = 0;
  for (int i = 0; i < n; i++) {
    res += sum(n) - sum(a[i]);
    ;
    add(a[i], n);
  }
  if (res & 1)
    cout << "NO" << endl;
  else
    cout << "YES" << endl;
}
int main() {
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```