#include <bits/stdc++.h>
using namespace std;
void _print(long long t) { cerr << t; }
void _print(string t) { cerr << t; }
void _print(char t) { cerr << t; }
void _print(long double t) { cerr << t; }
void _print(double t) { cerr << t; }
template <class T, class V>
void _print(pair<T, V> p);
template <class T>
void _print(vector<T> v);
template <class T>
void _print(set<T> v);
template <class T, class V>
void _print(map<T, V> v);
template <class T>
void _print(multiset<T> v);
template <class T, class V>
void _print(pair<T, V> p) {
  cerr << "{";
  _print(p.first);
  cerr << ",";
  _print(p.second);
  cerr << "}";
}
template <class T>
void _print(vector<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(set<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(multiset<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T, class V>
void _print(map<T, V> v) {
  cerr << "[ ";
  for (auto i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
long long mpow(long long base, long long exp);
void ipgraph(long long m);
void dfs(long long u, long long par);
const long long mod = 1000000007;
long long par[500000 + 5];
long long rnk[500000 + 5];
void make_set(long long n) {
  for (long long i = 1; i <= n; ++i) {
    par[i] = i;
    rnk[i] = 1;
  }
}
long long find_par(long long a) {
  if (par[a] == a) return a;
  return par[a] = find_par(par[a]);
}
void make_union(long long a, long long b) {
  long long p1 = find_par(a);
  long long p2 = find_par(b);
  if (p1 == p2) {
    return;
  }
  if (rnk[p1] > rnk[p2]) {
    par[p2] = p1;
    rnk[p1] += rnk[p2];
  } else {
    par[p1] = p2;
    rnk[p2] += rnk[p1];
  }
}
void sol() {
  long long n, i, j, k, m, q;
  bool pos = true;
  cin >> n;
  make_set(n);
  std::vector<long long> a(n, 0);
  std::vector<pair<long long, long long> > v, v_index;
  unordered_map<long long, long long> make_pair;
  unordered_set<long long> st;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    v.push_back({a[i], i + 1});
    st.insert(a[i]);
  }
  sort(v.begin(), v.end());
  std::vector<long long> b = a;
  sort(b.begin(), b.end());
  if (a == b) {
    cout << "YES" << '\n';
    return;
  }
  for (long long i = 0; i < n; i++) {
    make_union(v[i].second, i + 1);
  }
  std::vector<bool> done(n + 5, false);
  long long even = 0;
  for (long long i = 1; i <= n; i++) {
    long long temp = rnk[find_par(i)];
    if (find_par(i) == i) {
      if (temp % 2 == 0) {
        even++;
        pos = false;
      }
    }
  }
  if (st.size() == a.size() && even % 2 == 1) {
    cout << "NO" << '\n';
    return;
  }
  cout << "YES" << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) sol();
  return 0;
}
long long mpow(long long base, long long exp) {
  base %= mod;
  long long result = 1;
  while (exp > 0) {
    if (exp & 1) result = (result * base) % mod;
    base = (base * base) % mod;
    exp >>= 1;
  }
  return result;
}

### Prompt

Generate 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;
void _print(long long t) { cerr << t; }
void _print(string t) { cerr << t; }
void _print(char t) { cerr << t; }
void _print(long double t) { cerr << t; }
void _print(double t) { cerr << t; }
template <class T, class V>
void _print(pair<T, V> p);
template <class T>
void _print(vector<T> v);
template <class T>
void _print(set<T> v);
template <class T, class V>
void _print(map<T, V> v);
template <class T>
void _print(multiset<T> v);
template <class T, class V>
void _print(pair<T, V> p) {
  cerr << "{";
  _print(p.first);
  cerr << ",";
  _print(p.second);
  cerr << "}";
}
template <class T>
void _print(vector<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(set<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(multiset<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T, class V>
void _print(map<T, V> v) {
  cerr << "[ ";
  for (auto i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
long long mpow(long long base, long long exp);
void ipgraph(long long m);
void dfs(long long u, long long par);
const long long mod = 1000000007;
long long par[500000 + 5];
long long rnk[500000 + 5];
void make_set(long long n) {
  for (long long i = 1; i <= n; ++i) {
    par[i] = i;
    rnk[i] = 1;
  }
}
long long find_par(long long a) {
  if (par[a] == a) return a;
  return par[a] = find_par(par[a]);
}
void make_union(long long a, long long b) {
  long long p1 = find_par(a);
  long long p2 = find_par(b);
  if (p1 == p2) {
    return;
  }
  if (rnk[p1] > rnk[p2]) {
    par[p2] = p1;
    rnk[p1] += rnk[p2];
  } else {
    par[p1] = p2;
    rnk[p2] += rnk[p1];
  }
}
void sol() {
  long long n, i, j, k, m, q;
  bool pos = true;
  cin >> n;
  make_set(n);
  std::vector<long long> a(n, 0);
  std::vector<pair<long long, long long> > v, v_index;
  unordered_map<long long, long long> make_pair;
  unordered_set<long long> st;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    v.push_back({a[i], i + 1});
    st.insert(a[i]);
  }
  sort(v.begin(), v.end());
  std::vector<long long> b = a;
  sort(b.begin(), b.end());
  if (a == b) {
    cout << "YES" << '\n';
    return;
  }
  for (long long i = 0; i < n; i++) {
    make_union(v[i].second, i + 1);
  }
  std::vector<bool> done(n + 5, false);
  long long even = 0;
  for (long long i = 1; i <= n; i++) {
    long long temp = rnk[find_par(i)];
    if (find_par(i) == i) {
      if (temp % 2 == 0) {
        even++;
        pos = false;
      }
    }
  }
  if (st.size() == a.size() && even % 2 == 1) {
    cout << "NO" << '\n';
    return;
  }
  cout << "YES" << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) sol();
  return 0;
}
long long mpow(long long base, long long exp) {
  base %= mod;
  long long result = 1;
  while (exp > 0) {
    if (exp & 1) result = (result * base) % mod;
    base = (base * base) % mod;
    exp >>= 1;
  }
  return result;
}

```