#include <bits/stdc++.h>
using namespace std;
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
template <typename T_container, typename T = typename enable_if<
                                    !is_same<T_container, string>::value,
                                    typename T_container::value_type>::type>
ostream &operator<<(ostream &os, const T_container &v) {
  os << '{';
  string sep;
  for (const T &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
void dbg_out() { cerr << endl; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << ' ' << H;
  dbg_out(T...);
}
using ll = long long;
using ull = unsigned long long;
mt19937 rng(
    (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
const int mod = 998244353;
const int inf = 0x3f3f3f3f;
const int maxn = 2e5 + 10;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int _;
  cin >> _;
  while (_--) {
    int n;
    cin >> n;
    vector<int> a(n + 1), pos(n + 1);
    bool f = false;
    for (int i = int(1); i <= (int)(n); i++) {
      int x;
      cin >> x;
      if (pos[x] != 0) f = true;
      pos[x] = i;
      a[i] = x;
    }
    if (f) {
      cout << "YES" << '\n';
      continue;
    }
    int ans = 0;
    for (int i = int(1); i <= (int)(n); i++) {
      if (a[i] == i) continue;
      ans++;
      swap(pos[a[i]], pos[i]);
      swap(a[pos[a[i]]], a[pos[i]]);
    }
    if (ans % 2 == 0)
      cout << "YES" << '\n';
    else
      cout << "NO" << '\n';
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
template <typename T_container, typename T = typename enable_if<
                                    !is_same<T_container, string>::value,
                                    typename T_container::value_type>::type>
ostream &operator<<(ostream &os, const T_container &v) {
  os << '{';
  string sep;
  for (const T &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
void dbg_out() { cerr << endl; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << ' ' << H;
  dbg_out(T...);
}
using ll = long long;
using ull = unsigned long long;
mt19937 rng(
    (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
const int mod = 998244353;
const int inf = 0x3f3f3f3f;
const int maxn = 2e5 + 10;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int _;
  cin >> _;
  while (_--) {
    int n;
    cin >> n;
    vector<int> a(n + 1), pos(n + 1);
    bool f = false;
    for (int i = int(1); i <= (int)(n); i++) {
      int x;
      cin >> x;
      if (pos[x] != 0) f = true;
      pos[x] = i;
      a[i] = x;
    }
    if (f) {
      cout << "YES" << '\n';
      continue;
    }
    int ans = 0;
    for (int i = int(1); i <= (int)(n); i++) {
      if (a[i] == i) continue;
      ans++;
      swap(pos[a[i]], pos[i]);
      swap(a[pos[a[i]]], a[pos[i]]);
    }
    if (ans % 2 == 0)
      cout << "YES" << '\n';
    else
      cout << "NO" << '\n';
  }
  return 0;
}

```