#include <bits/stdc++.h>
using namespace std;
const int N = 5e5 + 5;
struct FT {
  vector<long long> s;
  FT(int n) : s(n) {}
  void upd(int pos, long long dif) {
    for (; pos < (int)(s).size(); pos |= pos + 1) s[pos] += dif;
  }
  long long qry(int pos) {
    long long res = 0;
    for (; pos > 0; pos &= pos - 1) res += s[pos - 1];
    return res;
  }
  int lower_bound(long long sum) {
    if (sum <= 0) return -1;
    int pos = 0;
    for (int pw = 1 << 25; pw; pw >>= 1) {
      if (pos + pw <= (int)(s).size() && s[pos + pw - 1] < sum)
        pos += pw, sum -= s[pos - 1];
    }
    return pos;
  }
};
int n, a[N];
bool vis[N];
void solve() {
  cin >> n;
  fill(vis, vis + n, 0);
  bool ok = false;
  for (int i = 0; i < (n); ++i) {
    cin >> a[i];
    --a[i];
    if (vis[a[i]]) {
      ok = true;
    } else {
      vis[a[i]] = true;
    }
  }
  FT ft(n);
  long long cnt = 0;
  for (int i = 0; i < (n); ++i) {
    cnt += ft.qry(n) - ft.qry(a[i] + 1);
    ft.upd(a[i], 1);
  }
  cout << (cnt % 2 == 0 || ok ? "YES\n" : "NO\n");
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(cin.failbit);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Your task is to create a Cpp solution 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;
struct FT {
  vector<long long> s;
  FT(int n) : s(n) {}
  void upd(int pos, long long dif) {
    for (; pos < (int)(s).size(); pos |= pos + 1) s[pos] += dif;
  }
  long long qry(int pos) {
    long long res = 0;
    for (; pos > 0; pos &= pos - 1) res += s[pos - 1];
    return res;
  }
  int lower_bound(long long sum) {
    if (sum <= 0) return -1;
    int pos = 0;
    for (int pw = 1 << 25; pw; pw >>= 1) {
      if (pos + pw <= (int)(s).size() && s[pos + pw - 1] < sum)
        pos += pw, sum -= s[pos - 1];
    }
    return pos;
  }
};
int n, a[N];
bool vis[N];
void solve() {
  cin >> n;
  fill(vis, vis + n, 0);
  bool ok = false;
  for (int i = 0; i < (n); ++i) {
    cin >> a[i];
    --a[i];
    if (vis[a[i]]) {
      ok = true;
    } else {
      vis[a[i]] = true;
    }
  }
  FT ft(n);
  long long cnt = 0;
  for (int i = 0; i < (n); ++i) {
    cnt += ft.qry(n) - ft.qry(a[i] + 1);
    ft.upd(a[i], 1);
  }
  cout << (cnt % 2 == 0 || ok ? "YES\n" : "NO\n");
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(cin.failbit);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

```