#include <bits/stdc++.h>
using namespace std;
const int N = 5e5 + 5;
int T, n;
int a[N], b[N];
long long tree[N];
void add(int x, int k) {
  for (; x <= n; x += x & -x) tree[x] += k;
}
long long query(int x) {
  long long ans = 0;
  for (; x; x -= x & -x) ans += tree[x];
  return ans;
}
int main(int argc, char *argv[]) {
  cin >> T;
  while (T--) {
    long long ans = 0;
    cin >> n;
    for (int i = 1; i <= n; i++) b[i] = 0;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      b[a[i]]++;
    }
    if (n < 3) {
      if (n == 1) puts("YES");
      if (n == 2 && a[1] <= a[2])
        puts("YES");
      else if (n == 2)
        puts("NO");
      continue;
    }
    bool f = 1;
    for (int i = 1; i <= n; i++) {
      if (b[i] >= 2) {
        f = 0;
        break;
      }
    }
    if (!f) {
      puts("YES");
      continue;
    }
    memset(tree, 0, sizeof(tree));
    for (int i = 1; i <= n; i++) {
      ans += query(n) - query(a[i] - 1);
      add(a[i], 1);
    }
    if (ans % 2)
      puts("NO");
    else
      puts("YES");
  }
  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>
using namespace std;
const int N = 5e5 + 5;
int T, n;
int a[N], b[N];
long long tree[N];
void add(int x, int k) {
  for (; x <= n; x += x & -x) tree[x] += k;
}
long long query(int x) {
  long long ans = 0;
  for (; x; x -= x & -x) ans += tree[x];
  return ans;
}
int main(int argc, char *argv[]) {
  cin >> T;
  while (T--) {
    long long ans = 0;
    cin >> n;
    for (int i = 1; i <= n; i++) b[i] = 0;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      b[a[i]]++;
    }
    if (n < 3) {
      if (n == 1) puts("YES");
      if (n == 2 && a[1] <= a[2])
        puts("YES");
      else if (n == 2)
        puts("NO");
      continue;
    }
    bool f = 1;
    for (int i = 1; i <= n; i++) {
      if (b[i] >= 2) {
        f = 0;
        break;
      }
    }
    if (!f) {
      puts("YES");
      continue;
    }
    memset(tree, 0, sizeof(tree));
    for (int i = 1; i <= n; i++) {
      ans += query(n) - query(a[i] - 1);
      add(a[i], 1);
    }
    if (ans % 2)
      puts("NO");
    else
      puts("YES");
  }
  return 0;
}

```